Trigonometry

🧭 Overview

🧠 One-sentence thesis

Understanding the relationships among angles and sides in triangles—especially right, isosceles, and equilateral triangles—and the properties of angles formed by parallel lines provides the geometric foundation for trigonometry.

📌 Key points (3–5)

• Sum of angles in any triangle: always equals 180°, which is the basis for finding unknown angles.
• Special triangle properties: right triangles have one 90° angle; equilateral triangles have all angles equal (60° each); isosceles triangles have two equal base angles.
• Angle relationships: vertical angles are equal; supplementary angles sum to 180°; complementary angles sum to 90°.
• Common confusion: vertex angle vs. vertical angles—vertex angle is the angle between the equal sides of an isosceles triangle; vertical angles are non-adjacent angles formed by intersecting lines.
• Parallel lines and transversals: when a line crosses two parallel lines, alternate interior angles and corresponding angles are equal.

📐 Fundamental angle measurements

📐 Basic angle units and types

One full rotation = 360°
A straight angle = 180° (half rotation)
A right angle = 90° (quarter rotation)

• Degrees are the standard unit for measuring angles.
• These reference values anchor all other angle calculations.

🔺 Acute vs. obtuse angles

• Acute angles: between 0° and 90°.
• Obtuse angles: between 90° and 180°.
• Example: In a triangle, you cannot have more than one obtuse angle because the sum of all three angles must equal 180°.

➕ Supplementary and complementary angles

Supplementary angles: two angles that add to 180°
Complementary angles: two angles that add to 90°

• Supplementary angles often appear when two angles form a straight line.
• In a right triangle, the two smaller angles are complementary (they sum to 90°).
• Don't confuse: supplementary (180°) vs. complementary (90°).

🔺 Core triangle properties

🔺 Sum of angles in a triangle

The sum of the angles in a triangle is 180°.

• This is the most fundamental triangle fact.
• If you tear off the three corners of any triangle and line them up, they form a straight angle (180°).
• Example: If two angles are 25° and 115°, the third angle x satisfies x + 25 + 115 = 180, so x = 40°.

📏 Relationship between sides and angles

• Standard labeling: angles are A, B, C; the side opposite angle A is called a, opposite B is b, opposite C is c.
• Key observation from the excerpt's activities: when sides are ordered a > b > c, the angles opposite them are ordered A > B > C.
• In other words, the largest angle is opposite the longest side, and the smallest angle is opposite the shortest side.

🔲 Special triangle types

⊿ Right triangles

A right triangle has one angle of 90°.

• The side opposite the 90° angle is the hypotenuse, which is always the longest side.
• The two smaller angles in a right triangle are complementary (they sum to 90°).
• Example: If one smaller angle is 34°, the other is 90 − 34 = 56°.

🔺 Pythagorean theorem and its converse

Pythagorean theorem: If a, b, and c are the sides of a right triangle and c is the hypotenuse, then a² + b² = c².

• The hypothesis is the "if" part; the conclusion is the "then" part.
• The converse interchanges hypothesis and conclusion: If a² + b² = c², then the triangle is a right triangle.
• The converse is also true and can be used to test whether a triangle is right.
• Example: For a = 12, b = 16, c = 20, check 12² + 16² = 144 + 256 = 400 = 20², so it is a right triangle.

🔼 Equilateral triangles

An equilateral triangle has all three sides the same length.
All of the angles of an equilateral triangle are equal.

• Since the sum of angles is 180° and all three are equal, each angle is 180 ÷ 3 = 60°.
• Example: If all three sides are 4 feet, then 3x = 180, so x = 60°.

🔻 Isosceles triangles

An isosceles triangle has two sides of equal length.
The base angles of an isosceles triangle are equal.

• The vertex angle is the angle between the two equal sides.
• The other two angles are the base angles, and they are equal.
• Example: If the base angles are each 38°, the vertex angle x satisfies x + 38 + 38 = 180, so x = 104°.
• Don't confuse: the vertex angle (part of the triangle) with vertical angles (formed by intersecting lines).

🔀 Angle relationships from intersecting and parallel lines

✖️ Vertical angles

Vertical angles: non-adjacent angles formed by the intersection of two straight lines.
Vertical angles are equal.

• When two lines cross, they form four angles; opposite (non-adjacent) pairs are vertical angles.
• Example: If one angle is 50°, the vertical angle opposite it is also 50°.

⫴ Parallel lines cut by a transversal

When parallel lines are intersected by a transversal, the alternate interior angles are equal. Corresponding angles are also equal.

• A transversal is a line that intersects two parallel lines, forming eight angles.
• Corresponding angles: angles in the same position relative to the transversal on each parallel line (e.g., angles 1 and 5 in the excerpt's diagram).
• Alternate interior angles: angles on opposite sides of the transversal between the parallel lines (e.g., angles 3 and 6, or 4 and 5).
• Example: In a parallelogram formed by two sets of parallel lines, opposite angles are equal because alternate interior and corresponding angles are equal.
• Adjacent angles of a parallelogram are supplementary (they sum to 180°).

📊 Summary of key geometric facts

Fact	Statement	Example use
1. Sum of angles in a triangle	Sum = 180°	Find the third angle when two are known
2. Right triangle	One angle = 90°	The two smaller angles are complementary
3. Equilateral triangle	All angles equal	Each angle = 60°
4. Isosceles triangle	Base angles equal	Find vertex angle from base angles
5. Vertical angles	Vertical angles equal	Find unknown angles at intersections
6. Parallel lines + transversal	Alternate interior and corresponding angles equal	Find angles in parallelograms and other figures

🧮 Exterior angles (from homework problems)

• An exterior angle of a triangle is formed by one side and the extension of an adjacent side.
• The excerpt's problems suggest finding a rule: the exterior angle equals the sum of the two non-adjacent interior angles (α and β in the diagram).
• Example: If the two non-adjacent interior angles are 40° and 80°, the exterior angle is 40 + 80 = 120°.

🔤 Greek letters for angles

• Trigonometry often uses lowercase Greek letters to denote angle measures: α (alpha), β (beta), γ (gamma), θ (theta), φ (phi), ψ (psi), etc.
• The excerpt provides a full table of the Greek alphabet for reference.

🧭 Overview

🧠 One-sentence thesis

Congruent triangles have identical size and shape with all corresponding angles and sides equal, and the altitude of an equilateral triangle divides it into two congruent right triangles.

📌 Key points (3–5)

• What congruence means: two triangles are congruent when they have exactly the same size and shape—all corresponding angles are equal and all corresponding sides have the same lengths.
• How to identify corresponding parts: match angles and sides that occupy the same position in each triangle; equal angles are opposite equal sides.
• Common confusion: congruence requires both equal angles and equal side lengths; equal angles alone do not guarantee congruence.
• Altitude property: the altitude of an equilateral triangle (perpendicular from a vertex to the opposite side) creates two congruent right triangles.

🔍 Definition and meaning

🔍 What congruent triangles are

Congruent triangles: two triangles that have exactly the same size and shape.

• This definition has two requirements:
  • Corresponding angles are equal: each angle in one triangle equals the matching angle in the other.
  • Corresponding sides have the same lengths: each side in one triangle equals the matching side in the other.
• "Corresponding" means the parts that occupy the same relative position in each triangle.
• Example: if triangle ABC is congruent to triangle DEF, then angle A = angle D, angle B = angle E, angle C = angle F, and side a = side d, side b = side e, side c = side f.

🧩 How to find corresponding parts

• Match by position: angles and sides that play the same role in each triangle correspond to each other.
• Use known equalities: if two angles are already known to be equal (e.g., both are right angles or vertical angles), they are corresponding angles.
• Opposite relationships: the side opposite a given angle in one triangle corresponds to the side opposite the equal angle in the other triangle.
• Don't confuse: corresponding parts are determined by the triangle's structure, not by visual proximity or orientation on the page.

📐 Working with congruent triangles

📐 Finding unknown angles and sides

When two triangles are congruent, you can use the equality of corresponding parts to find unknowns.

Steps:

1. Identify which angles or sides are already known to be equal (e.g., right angles, vertical angles, given measurements).
2. Use the fact that the sum of angles in a triangle is 180° to find any remaining angles.
3. Match corresponding sides: if two sides are known to be equal, the sides opposite equal angles are also equal.

Example: In the excerpt, triangle ABC and triangle DCE are congruent. Angle B = angle D (both 90°), and angle BCA = angle DCE (vertical angles, both 25°). Therefore angle A = angle E = 65° (since 180° - 90° - 25° = 65°). Side AB = side DE, side BC = side CD, and side AC = side CE, so z = 9.

🔺 Special case: equilateral triangles and altitudes

Altitude of a triangle: the segment from one vertex perpendicular to the opposite side.

• In an equilateral triangle, all three sides are equal and all three angles are 60°.
• When you draw the altitude from one vertex to the opposite side, it creates a right angle (90°) with that side.
• This altitude divides the equilateral triangle into two congruent right triangles.
• Why they are congruent:
  • Both have a right angle (90°).
  • Both share the altitude as a common side.
  • The original 60° angle at the top is split into two 30° angles (one in each right triangle).
  • The base is divided into two equal segments (because the altitude bisects the base in an equilateral triangle).
• Example: An equilateral triangle with sides of 8 units is divided by its altitude into two right triangles, each with a 30° angle, a 60° angle, a 90° angle, a base of 4 units, and the altitude as the shared side.

⚠️ Common confusions

⚠️ Congruence vs. equal angles only

• Congruence requires both: equal angles and equal side lengths.
• Equal angles alone mean the triangles are similar in shape but not necessarily the same size.
• Equal side lengths alone (without knowing angles) also do not guarantee congruence unless you know the triangles are the same type (e.g., both equilateral).
• Don't confuse: "corresponding angles are equal" is a necessary condition for congruence, but not sufficient by itself.

⚠️ Corresponding vs. adjacent

• Corresponding parts are determined by the triangle's internal structure (which angle matches which), not by how the triangles are drawn next to each other.
• Example: In the excerpt, angle BCA and angle DCE are corresponding because they are vertical angles (formed by intersecting lines), not because they are next to each other in the diagram.

🧭 Overview

🧠 One-sentence thesis

The distance formula, derived from the Pythagorean theorem applied to map coordinates, allows us to calculate the straight-line distance between any two points by treating their coordinate differences as the legs of a right triangle.

📌 Key points (3–5)

• What the distance formula does: calculates the straight-line distance between two points using their coordinates.
• How it connects to geometry: the formula applies the Pythagorean theorem to a right triangle formed by horizontal and vertical coordinate differences.
• Practical use case: measuring distances on maps when coordinates are known (e.g., trail distances in hiking).
• Common confusion: the excerpt introduces the setup but does not yet show the complete formula—don't assume the method is finished; the key is recognizing that coordinate differences become triangle sides.

📐 The Pythagorean theorem foundation

📐 What the Pythagorean theorem is

The Pythagorean theorem: in a right triangle, the relationship between the lengths of the sides allows calculation of the hypotenuse (or any side) from the other two sides.

• The excerpt does not state the formula explicitly, but it emphasizes that Delbert "remembers the Pythagorean theorem" and uses it with map coordinates.
• The theorem is the mathematical backbone for finding distances when you cannot measure directly.

🗺️ Applying coordinates to the theorem

• Delbert's map uses a grid where each interval represents one kilometer.
• The two locations are at different grid positions: Sycamore Canyon trail head at 12-C and Coyote Trail junction at 8-F.
• By treating the horizontal and vertical differences between these coordinates as the two legs of a right triangle, the straight-line distance becomes the hypotenuse.
• Example: if one point is 4 units east and 3 units north of another, those differences (4 and 3) are the legs; the Pythagorean theorem gives the hypotenuse (straight-line distance).

🧭 Setting up the distance calculation

🧭 Identifying the coordinate differences

• The excerpt states that Delbert "uses the map coordinates to label the sides of" a right triangle (the sentence is incomplete, but the intent is clear).
• The horizontal difference and vertical difference between the two points become the two legs of the triangle.
• Don't confuse: the coordinates themselves (e.g., "12-C" and "8-F") are not the side lengths—you must subtract corresponding coordinates to find the differences.

📏 Why this method works for maps

• Maps with uniform grids allow each coordinate difference to represent a real distance (here, one kilometer per interval).
• The straight-line distance (as the crow flies) is not the same as walking distance along trails, but it is useful for planning and estimation.
• Example: Delbert cannot walk directly across terrain, but knowing the straight-line distance helps him estimate travel time and compare routes.

🔍 What the excerpt does not yet show

🔍 Incomplete presentation

• The excerpt introduces the scenario and mentions the Pythagorean theorem, but it cuts off before showing the actual calculation or the distance formula itself.
• The full distance formula (typically written as the square root of the sum of squared coordinate differences) is not stated here.
• For self-study: recognize that this section is setting up the concept; the formula and worked example likely appear in the continuation of the text.

🧭 Overview

🧠 One-sentence thesis

Congruent triangles have identical size and shape with equal corresponding angles and sides, while similar triangles share the same shape but differ in size with equal corresponding angles and proportional corresponding sides.

📌 Key points (3–5)

• Congruent triangles: exactly the same size and shape; all corresponding angles equal and all corresponding sides equal in length.
• Similar triangles: same shape but not necessarily same size; corresponding angles equal and corresponding sides proportional.
• Shortcut for similarity: you only need to verify one condition (either all angles equal OR all sides proportional) and the other follows automatically.
• Common confusion: congruent vs similar—congruent means identical in every way; similar means same shape but can be scaled up or down.
• Practical use: proportions from similar triangles allow indirect measurement of unknown lengths.

🔷 Congruent Triangles

🔷 What congruence means

Two triangles are congruent if they have exactly the same size and shape.

• This means two things must both be true:
  • Their corresponding angles are equal.
  • Their corresponding sides have the same lengths.
• Think of congruent triangles as perfect copies of each other.

🔍 Finding unknown parts

When two triangles are congruent, you can find missing angles and sides by matching corresponding parts.

Example: If triangle ABC is congruent to triangle DEF, and angle B = angle D (both right angles), and angle BCA = angle DCE (vertical angles), then the third angles must also match. If one triangle has a 25° angle, the corresponding angle in the other triangle is also 25°.

📐 Special case: equilateral triangles

The altitude of an equilateral triangle divides it into two congruent right triangles.

• Each original angle is 60°.
• The altitude creates two right triangles, each with angles 30°, 60°, and 90°.
• These are called 30°–60°–90° triangles.
• Important property: the leg opposite the 30° angle is half the length of the hypotenuse.

🔶 Similar Triangles

🔶 What similarity means

Two triangles are similar if they have the same shape but not necessarily the same size.

• Corresponding angles are equal.
• Corresponding sides are proportional (not equal, but in the same ratio).
• One similar triangle is like an enlargement or reduction of the other.

⚡ Shortcut for checking similarity

You don't need to check both conditions:

What to verify	Result
All corresponding angles are equal	Sides will automatically be proportional
All corresponding sides are proportional	Angles will automatically be equal

Don't confuse: You need all three angle pairs equal (or all three side ratios equal), not just one or two.

🎯 Even shorter shortcut for right triangles

If two right triangles have one pair of corresponding acute angles equal, the triangles are similar.

• You already know one angle pair is equal (both 90°).
• If one more acute angle matches, the third angle must also match (since angles sum to 180°).

📏 Using Proportions with Similar Triangles

📏 Setting up proportions

When triangles are similar, corresponding sides are proportional:

• Write ratios in consistent order: larger triangle over smaller triangle (or vice versa, but stay consistent).
• Match corresponding parts: shortest to shortest, longest to longest, etc.

Example: If the shorter legs are 9 and 6, longer legs are 12 and 8, and hypotenuses are 15 and 10:

• 9/6 = 12/8 = 15/10 = 1.5
• All ratios equal means the triangles are similar.

🔧 Solving for unknown sides

To find an unknown side in a similar triangle:

1. Identify which sides correspond.
2. Set up a proportion using known sides.
3. Cross-multiply to solve.

Example: If x/4 = 15/6, then cross-multiply: 6x = 4(15) = 60, so x = 10.

🌍 Indirect measurement

Similar triangles allow you to measure things that are hard to reach directly.

Example: To find a building's height:

• A 5-foot pole casts a 3-foot shadow.
• The building casts a 12-foot shadow.
• Set up proportion: h/12 = 5/3 (where h is building height).
• The sun's rays are parallel, so the angles at the shadow tips are equal, making the triangles similar.

🧭 Overview

🧠 One-sentence thesis

By using trigonometric ratios (sine, cosine, and tangent), we can find unknown sides of a right triangle when we know only one side and one acute angle, because these ratios remain constant for any given angle across all similar triangles.

📌 Key points (3–5)

• What trig ratios measure: sine, cosine, and tangent are ratios between specific sides of a right triangle that depend only on the angle, not the triangle's size.
• Why ratios are constant: similar triangles with the same angle always have the same side ratios, making these values predictable and useful.
• How to choose the right ratio: sine relates opposite side to hypotenuse; cosine relates adjacent side to hypotenuse; tangent relates opposite to adjacent.
• Common confusion: these definitions apply only to right triangles; you must identify which side is opposite, adjacent, or hypotenuse relative to the angle in question.
• Practical power: knowing one side and one acute angle is enough to solve for all other sides, unlike the Pythagorean theorem which requires two sides.

📐 The three fundamental ratios

📐 Sine of an angle

Sine of an acute angle: sin(θ) = opposite / hypotenuse

• The sine is the ratio of the side opposite the angle to the hypotenuse.
• This ratio is the same for every right triangle containing that angle, because all such triangles are similar.
• Example: In any 30° right triangle, sin(30°) = 0.5, meaning the opposite side is always half the hypotenuse length.
• Don't confuse: You must use a right triangle; the ratio doesn't work for other triangle types.

📐 Cosine of an angle

Cosine of an acute angle: cos(θ) = adjacent / hypotenuse

• The cosine is the ratio of the side adjacent to the angle to the hypotenuse.
• "Adjacent" means the side that touches the angle (but is not the hypotenuse).
• As the angle increases from 0° to 90°, cosine decreases from 1 to 0, because the adjacent side gets shorter while the hypotenuse stays the same.

📐 Tangent of an angle

Tangent of an acute angle: tan(θ) = opposite / adjacent

• The tangent is the ratio of the opposite side to the adjacent side.
• Unlike sine and cosine, tangent does not involve the hypotenuse.
• Example: To find the height of a cliff when you know the distance to its base and the angle of elevation, use tangent because you have adjacent and want opposite.
• As the angle increases toward 90°, tangent grows without bound (the opposite side gets much longer than the adjacent side).

🔢 Using calculators and tables

🔢 Finding trig values

• Mathematicians have calculated sine, cosine, and tangent for every angle; these values are stored in scientific calculators.
• Example: sin(50°) = 0.7660 (rounded to four decimal places); the full value is irrational and continues indefinitely.
• Calculator note: When you press SIN, the calculator shows "sin(" with an open parenthesis, prompting you to enter the angle, because sine is a function of the angle.

🔢 Accuracy in calculations

• Most trig values are irrational numbers, so calculators show as many digits as the display allows.
• Don't round intermediate steps: rounding too early loses accuracy; use the calculator's stored value throughout the calculation, then round the final answer.
• Example keystrokes to avoid rounding: sin(50) × 18 (rather than entering 0.7660 × 18).

🧮 Solving for unknown sides

🧮 Setting up the equation

1. Identify which side you know and which you want to find.
2. Choose the trig ratio that connects those two sides.
3. Write the ratio equation: trig(angle) = (side you want) / (side you know), or the reverse.
4. Solve for the unknown side.

🧮 Example: finding the opposite side

• Given: hypotenuse = 18 cm, angle = 50°, find the opposite side x.
• Use sine because it relates opposite to hypotenuse: sin(50°) = x / 18.
• Calculate: 0.7660 = x / 18, so x = 18 × 0.7660 = 13.79 cm.

🧮 Example: finding the hypotenuse

• Given: opposite side = 7.5 m, angle = 62°, find hypotenuse x.
• Use sine: sin(62°) = 7.5 / x.
• Rearrange: x = 7.5 / sin(62°) ≈ 8.5 m.

🧮 Choosing the right ratio

You know	You want	Use this ratio
Hypotenuse + angle	Opposite side	Sine
Hypotenuse + angle	Adjacent side	Cosine
One leg + angle	Other leg	Tangent
Opposite + angle	Hypotenuse	Sine (rearranged)
Adjacent + angle	Hypotenuse	Cosine (rearranged)

🔗 Relationships among the ratios

🔗 Complementary angles

• In a right triangle, the two non-right angles add to 90°; they are complementary.
• Key pattern: sin(θ) = cos(90° - θ) and cos(θ) = sin(90° - θ).
• Why: the side opposite one angle is adjacent to the other angle, so their sine and cosine swap.
• Example: sin(30°) = cos(60°) = 0.5.

🔗 Connecting tangent to sine and cosine

• There is a relationship: tan(θ) = sin(θ) / cos(θ).
• This follows from the definitions: (opposite/hypotenuse) / (adjacent/hypotenuse) = opposite/adjacent.
• You can verify this by calculating all three ratios for any right triangle.

🔗 Behavior as angles change

• As θ increases from 0° to 90°:
  • sin(θ) increases from 0 to 1 (opposite side grows).
  • cos(θ) decreases from 1 to 0 (adjacent side shrinks).
  • tan(θ) increases from 0 toward infinity (opposite grows much faster than adjacent shrinks).
• Don't confuse: The graph of these values against angle is not a straight line; the rate of change varies.

⚠️ Common pitfalls and reminders

⚠️ Right triangles only

• These three trig ratio definitions apply only to right triangles.
• If given a non-right triangle, you may need to draw an altitude or other line to create a right triangle.
• Example: To find the altitude of an isosceles triangle, draw a line from the vertex angle perpendicular to the base, creating two congruent right triangles.

⚠️ Identifying sides correctly

• "Opposite" and "adjacent" are always relative to the angle you're considering.
• The hypotenuse is always the longest side, opposite the right angle.
• Don't confuse: In the same triangle, a side that is "opposite" one acute angle is "adjacent" to the other acute angle.

⚠️ Why ratios work

• The power of trig ratios comes from similar triangles: all right triangles with the same acute angle have the same shape, just different sizes.
• The ratio opposite/hypotenuse (for example) is identical whether the hypotenuse is 2 cm or 200 cm, as long as the angle is the same.
• This constancy is what makes it possible to tabulate and use these values universally.

🧭 Overview

🧠 One-sentence thesis

If we know one side and any other part of a right triangle (besides the right angle itself), we can use trigonometry to find all remaining unknown parts—a process called solving the triangle.

📌 Key points (3–5)

• What "solving" means: finding all six parts (three sides and three angles) of a right triangle when some are given.
• Minimum information needed: one side plus one other part (angle or side) beyond the right angle.
• Inverse trig functions: sin⁻¹, cos⁻¹, and tan⁻¹ let us find an angle when we know its trigonometric ratio.
• Common confusion: sin⁻¹(x) does NOT mean 1/sin(x); it means "the angle whose sine is x."
• Special angles (30°, 45°, 60°): these have exact (non-decimal) trig ratios that should be memorized using two reference triangles.

🔧 The solving process

🔧 What you need to start

Solving a triangle: finding all unknown sides and angles when given enough information.

• For a right triangle, you already know one angle is 90°.
• You need at least one side plus one other part (another side or an acute angle).
• Use the appropriate trig ratio (sine, cosine, or tangent) based on which sides/angles you know and which you're finding.

🔧 Why use given values, not calculated ones

• The excerpt recommends using original given information rather than values you've already calculated.
• Reason: calculated values may carry rounding errors; using fresh given data avoids compounding mistakes.
• Example: If you know the hypotenuse and one angle, use sine to find the opposite side and cosine to find the adjacent side—don't find one side then use Pythagorean theorem with that rounded result.

🔄 Inverse trigonometric functions

🔄 What inverse trig does

• The tan⁻¹, sin⁻¹, and cos⁻¹ buttons find an angle when you know its trig ratio.
• Read as "inverse tangent," "inverse sine," "inverse cosine."
• Example: if tan(θ) = 0.5662, then θ = tan⁻¹(0.5662) ≈ 29.52°.

🔄 Two ways to say the same thing

The excerpt shows that these two statements are equivalent:

• sin(43.11°) = 0.6834
• sin⁻¹(0.6834) = 43.11°

Both mean "43.11° is the angle whose sine is 0.6834."

⚠️ Critical distinction: inverse vs reciprocal

Don't confuse:

• sin⁻¹(x) = "the angle whose sine is x"
• 1/sin(x) = "the reciprocal of the sine of angle x"

The notation sin⁻¹ does NOT mean "sine to the power of negative one" in the algebraic sense. Although a⁻¹ = 1/a for numbers, "sin" is not a variable—it's a function name, so sin⁻¹ denotes the inverse function.

Example: If you want the reciprocal of sin(x), write it as 1/sin(x), not sin⁻¹(x).

⭐ The special angles

⭐ Which angles are special and why

The angles 30°, 45°, and 60° are called special because their trig ratios can be expressed as exact values (using square roots and fractions), not just decimal approximations.

Angle	Sine	Cosine	Tangent
30°	1/2	√3/2	1/√3
45°	1/√2	1/√2	1
60°	√3/2	1/2	√3

⭐ Two reference triangles to memorize

The excerpt recommends remembering these exact values by knowing two triangles:

1. 45-45-90 triangle: legs of length 1, hypotenuse √2
2. 30-60-90 triangle: short leg 1, long leg √3, hypotenuse 2

From these triangles, you can always reconstruct the three trig ratios for each special angle.

⭐ Exact vs approximate values

• Exact values (like √3/2) are the true, non-rounded answers.
• Decimal approximations (like 0.8660) are rounded and not perfectly accurate, even if your calculator shows many digits.
• For most practical calculations, approximations are fine, but exact values let you find answers to any desired precision.

Example: The altitude of an equilateral triangle with 8 cm sides is exactly 4√3 cm, which is approximately 6.9282 cm. The exact answer is preferred because you can round it to any accuracy you need later.

⭐ Using special angles as benchmarks

• You can use the known values to estimate unknown angles mentally.
• Example: If sin(θ) = 0.95, and you know sin(60°) ≈ 0.8660, then θ must be greater than 60° (because sine increases as the angle increases from 0° to 90°).
• Similarly, cosine decreases as the angle increases from 0° to 90°, so if cos(α) > √3/2 (which equals cos(30°)), then α must be less than 30°.

📐 Worked scenarios

📐 Finding sides from an angle

Example: A 10-foot ladder leans against a wall at a 75° angle with the ground.

• To find the base distance (adjacent side): use cos(75°) = adjacent/hypotenuse → adjacent = 10 × cos(75°) ≈ 2.6 feet.
• To find the height up the wall (opposite side): use sin(75°) = opposite/hypotenuse → opposite = 10 × sin(75°) ≈ 9.7 feet.

📐 Finding an angle from sides

Example: A slide is 77 inches high and covers 136 inches horizontally.

• The slope is 77/136 = 0.5662, which is also tan(θ).
• Use inverse tangent: θ = tan⁻¹(0.5662) ≈ 29.52°.

📐 Using exact values in context

Example: A triangular prism box has equilateral triangle ends with 8 cm sides. Find the altitude.

• The altitude divides the triangle into two 30-60-90 triangles.
• The altitude is adjacent to the 30° angle; hypotenuse is 8 cm.
• cos(30°) = adjacent/hypotenuse → √3/2 = h/8 → h = 4√3 cm (exact) ≈ 6.93 cm (approximate).

📋 Summary concepts

📋 Key vocabulary

• Solve a triangle: find all unknown parts
• Inverse sine/cosine/tangent: find an angle from a ratio
• Special angles: 30°, 45°, 60°
• Exact value: non-rounded answer (often with radicals)
• Decimal approximation: rounded calculator output

📋 Study questions from the excerpt

1. How many parts do you need to know? At least three (including the right angle), with at least one being a side.
2. Why use given values? To avoid compounding rounding errors from calculated values.
3. What is the inverse trig button for? Finding an angle when you know its trig ratio.
4. Which are the special angles and why? 30°, 45°, 60°—because their trig ratios have exact radical/fractional forms.

🧭 Overview

🧠 One-sentence thesis

Trigonometric ratios extend to obtuse angles (90° to 180°) using coordinate definitions, enabling us to solve oblique triangles and compute areas when angles are not right angles.

📌 Key points (3–5)

• Coordinate definitions: For angles in standard position, cos(θ) = x/r, sin(θ) = y/r, tan(θ) = y/x, where (x, y) is any point on the terminal side and r is the distance from the origin.
• Supplementary angle identities: sin(180° − θ) = sin(θ), but cos(180° − θ) = −cos(θ) and tan(180° − θ) = −tan(θ).
• Common confusion: Two angles between 0° and 180° have the same sine (they are supplements); calculators return only one, so you must find the other manually.
• Negative ratios in quadrant II: Because x-coordinates are negative in the second quadrant, cosine and tangent are negative for obtuse angles, but sine remains positive.
• Area formula: For any triangle with sides a and b and included angle θ, Area = ½ab sin(θ).

📐 Standard position and coordinate definitions

📐 What standard position means

Standard position: An angle θ is placed with its vertex at the origin, initial side on the positive x-axis, and terminal side opening counter-clockwise.

• The initial side lies along the positive x-axis.
• The terminal side rotates counter-clockwise from the initial side.
• A point P(x, y) on the terminal side forms a right triangle by dropping a vertical line to the x-axis.
• The distance r from the origin to P is computed using the distance formula: r = √(x² + y²).

🔢 Coordinate definitions of trig ratios

Coordinate definitions: cos(θ) = x/r, sin(θ) = y/r, tan(θ) = y/x

• These definitions replace "adjacent/hypotenuse" and "opposite/hypotenuse" from right-triangle definitions.
• Why they work for any angle: The ratios depend only on coordinates, not on being "inside" a triangle.
• Independence of point choice: Any point on the terminal side gives the same ratios because similar triangles have proportional sides.
• Example: If P(12, 5) is on the terminal side, then r = 13, so cos(θ) = 12/13, sin(θ) = 5/13, tan(θ) = 5/12.

🔄 Why the same ratios from different points

• If you choose P(12, 5) or P'(24, 10) on the same terminal side, both yield identical trig ratios.
• The triangles formed are similar, so corresponding side ratios are equal.
• Example: For P'(24, 10), r = 26, giving cos(θ) = 24/26 = 12/13 (same as before).

🔺 Obtuse angles in the second quadrant

🔺 What makes obtuse angles different

Obtuse angle: An angle between 90° and 180°.

• The terminal side lies in the second quadrant (where x < 0, y > 0).
• Key consequence: x-coordinates are negative, so cos(θ) and tan(θ) are negative, but sin(θ) remains positive.
• Example: For θ with terminal side through (−4, 3), we have r = 5, so cos(θ) = −4/5, sin(θ) = 3/5, tan(θ) = −3/4.

🔁 Supplementary angle relationships

The excerpt establishes three identities for supplementary angles:

Identity	Meaning
sin(180° − θ) = sin(θ)	Supplements have equal sines
cos(180° − θ) = −cos(θ)	Supplements have opposite cosines
tan(180° − θ) = −tan(θ)	Supplements have opposite tangents

• Why sines are equal: Points (x, y) and (−x, y) have the same y-coordinate and same r, so y/r is identical.
• Why cosines differ in sign: The x-coordinates have opposite signs, so x/r values are negatives of each other.
• Example: sin(130°) = sin(50°) ≈ 0.7660, but cos(130°) = −cos(50°) ≈ −0.6428.

⚠️ Two angles with the same sine

• Common confusion: Between 0° and 180°, exactly two angles share the same sine—they are supplements.
• Your calculator's inverse sine (sin⁻¹) returns only the acute angle.
• How to find the obtuse solution: If sin(θ) = k and the calculator gives θ₁, the obtuse angle is θ₂ = 180° − θ₁.
• Example: sin(θ) = 0.25 → calculator gives 14.5°, so the obtuse solution is 180° − 14.5° = 165.5°.

🧮 Using calculators for obtuse angles

• Best approach: Use cosine when finding an obtuse angle, because cos⁻¹ of a negative value directly returns the obtuse angle.
• Example: If cos(θ) = −3/5, then cos⁻¹(−3/5) ≈ 126.9° (obtuse).
• Don't confuse: Using tan⁻¹ for an obtuse angle often returns a negative acute angle (the calculator's default range), not the obtuse angle you want.
• Example: If tan(θ) = −4/3, the calculator gives tan⁻¹(−4/3) ≈ −53.1°, but the obtuse angle is actually 180° − 53.1° = 126.9°.

🌟 Special angles in the second quadrant

🌟 Supplements of 30°, 45°, 60°

The supplements of the familiar special angles are 150°, 135°, and 120°.

Angle	cos(θ)	sin(θ)	tan(θ)
120°	−1/2	√3/2	−√3
135°	−1/√2	1/√2	−1
150°	−√3/2	1/2	−1/√3

• How to find them: Sketch the angle in standard position; the terminal side makes a 30°, 45°, or 60° angle with the negative x-axis.
• Example: For 135°, the terminal side passes through (−1, 1), so r = √2, giving cos(135°) = −1/√2, sin(135°) = 1/√2, tan(135°) = −1.

📏 Quadrantal angles

Quadrantal angles: Angles whose terminal sides lie on an axis (0°, 90°, 180°, 270°).

• For 90°: terminal side is the positive y-axis, so P(0, 1) gives cos(90°) = 0, sin(90°) = 1, tan(90°) is undefined (division by zero).
• For 180°: terminal side is the negative x-axis, so P(−1, 0) gives cos(180°) = −1, sin(180°) = 0, tan(180°) = 0.

📐 Area of a triangle using sine

📐 The area formula

Area formula: If a triangle has sides a and b with included angle θ, then Area = ½ab sin(θ).

• Why it works: The height h of the triangle satisfies sin(θ) = h/a, so h = a sin(θ); substituting into Area = ½ base × height gives ½b(a sin(θ)).
• This formula applies whether θ is acute, obtuse, or 90°.
• Example: A triangle with sides 6 and 7 and included angle 150° has area = ½(6)(7) sin(150°) = 21 sin(150°) = 21(1/2) = 21/2.

🏘️ Practical application: irregular lots

• Real-world use: Finding the area of an irregular quadrilateral (e.g., a land lot) by dividing it into two triangles.
• You need two sides and the included angle for each triangle.
• Example: Lot 86 is divided into two triangles; compute each area using the formula, then add them.
  • Lower triangle: a = 120.3, b = 141, θ = 95° → Area ≈ 8448.88 sq ft
  • Upper triangle: a = 161, b = 114.8, θ = 86.1° → Area ≈ 9220.00 sq ft
  • Total ≈ 17,669 sq ft

⚠️ Which sides to use

• Don't confuse: The formula uses the two sides that include (form) the known angle, not just any two sides.
• Example: In a triangle with sides labeled 5, b, c and angle φ between sides 5 and c, the area is ½(5)(c) sin(φ), not ½(5)(b) sin(φ).

🧭 Overview

🧠 One-sentence thesis

The Law of Sines extends trigonometry beyond right triangles by relating each side of any triangle to the sine of its opposite angle, enabling us to solve oblique triangles when we know certain combinations of sides and angles.

📌 Key points (3–5)

• What the Law of Sines states: In any triangle, the ratio of each side to the sine of its opposite angle is constant: sin(A)/a = sin(B)/b = sin(C)/c.
• When to use it for finding sides: You must know one side-angle pair (a side and its opposite angle) plus one other angle to find the side opposite that angle.
• When to use it for finding angles: You must know two sides and the angle opposite one of them; solve for the sine of the unknown angle.
• Common confusion—the ambiguous case: When finding an angle from its sine, two angles are possible (one acute, one obtuse); you must check whether both produce valid triangles.
• Why it matters: The Law of Sines solves real-world problems involving oblique triangles, from measuring inaccessible distances to calculating astronomical parallax.

📐 Deriving the Law of Sines

📐 The key technique: reducing to right triangles

The excerpt emphasizes a fundamental mathematical strategy:

"Reducing a new problem to an earlier one is a frequently used technique in mathematics."

• Start with an oblique (non-right) triangle ABC.
• Draw an altitude h from one vertex to the opposite side, creating two right triangles.
• In the two right triangles formed, write expressions for h using the sine ratio.

🔍 How the derivation works

Looking at triangle BCD: h/a = sin(C), so h = a·sin(C). Looking at triangle ABD: h/c = sin(A), so h = c·sin(A).

• Since both expressions equal h, set them equal: a·sin(C) = c·sin(A).
• Divide both sides by ac to get: sin(C)/c = sin(A)/a.
• By drawing a different altitude, you can similarly show sin(A)/a = sin(B)/b.

📋 The complete Law of Sines

Law of Sines: If the angles of a triangle are A, B, and C, and the opposite sides are respectively a, b, and c, then sin(A)/a = sin(B)/b = sin(C)/c, or equivalently a/sin(A) = b/sin(B) = c/sin(C).

• This law works for any triangle: acute, right, or obtuse.
• Each side is labeled with the lowercase letter corresponding to its opposite angle's uppercase letter.

🔧 Using the Law to Find Sides

🔧 What you need to know

To find an unknown side, you must have:

• One complete side-angle pair (a side and the angle opposite it).
• The angle opposite the unknown side.

🌊 Example scenario: measuring a ship's distance

Two observers 400 yards apart measure angles to a ship: one measures 83.2° from the shoreline, the other 79.4°.

• The unknown distance d is opposite the 79.4° angle.
• The known side is 400 yards; find its opposite angle: 180° − (79.4° + 83.2°) = 17.4°.
• Apply the Law: d/sin(79.4°) = 400/sin(17.4°).
• Solve: d = 400·sin(79.4°)/sin(17.4°) ≈ 1315 yards.

Don't confuse: You cannot use the Law of Sines if you only know two sides and an angle that is not opposite one of them—you need the opposite angle-side relationship.

🔍 Using the Law to Find Angles

🔍 What you need to know

To find an unknown angle, you must have:

• Two sides of the triangle.
• The angle opposite one of those sides.

⚠️ The ambiguous case

When you solve for sin(A) and get a value like 0.3723, two angles have that sine:

• The acute angle: A = sin⁻¹(0.3723) ≈ 21.9°.
• The obtuse angle: 180° − 21.9° = 158.1°.

How to decide which angle works:

• Check whether each angle fits in the triangle (all three angles must sum to 180°).
• Sometimes only the acute angle works.
• Sometimes only the obtuse angle works.
• Sometimes both produce valid but different triangles.

🔀 When two solutions exist

Example: B = 14.4°, a = 8, b = 3.

• Solving gives sin(A) ≈ 0.6632.
• Acute solution: A = 41.5°, which gives C = 124.1°.
• Obtuse solution: A = 138.5°, which gives C = 27.1°.
• Both triangles are valid; the problem has two distinct solutions.

Don't confuse: This ambiguity only occurs when you know two sides and an angle opposite one of them (SSA configuration), not when finding sides.

🌍 Real-World Applications

🏰 Multi-triangle problems

Sometimes you must solve two triangles to find the answer.

Example: Measuring a castle's height when you can't get close.

• First measurement: angle of elevation 18.5° from closest safe point.
• Second measurement: angle of elevation 15.9° from 20 yards farther back.
• Use the Law of Sines on the oblique triangle formed by the two observation points to find the distance r to the castle base.
• Then use the right triangle and sine ratio to find height h = r·sin(15.9°) ≈ 38.33 yards.

🌌 Astronomical parallax

Parallax: the apparent change in an object's position when viewed from two different locations.

• Two observers at known distance apart measure the angle to a distant object (like a star).
• The difference in viewing angles is the parallax angle p.
• The object and two observers form an (often nearly isosceles) triangle.
• Use the Law of Sines to calculate the distance.

Example: Asteroid with parallax 0.001° observed by astronomers 1000 km apart.

• The triangle has base 1000 km and base angles of 89.999° each.
• Distance x = 1000·sin(89.999°)/sin(0.001°) ≈ 57,000,000 km.

📏 Minutes and seconds for tiny angles

For very small angles (like stellar parallax), degrees are divided into smaller units:

Unit	Definition	Decimal equivalent
1 minute (1′)	1/60 of a degree	1° / 60
1 second (1″)	1/60 of a minute	1° / 3600

• Astronomical Unit (AU): the distance from Earth to the Sun, about 93 million miles or 1.5 × 10⁸ km.
• Parsec: the distance at which parallax from observations 1 AU apart equals 1″ (approximately 206,265 AU).

Example: Star Wolf 359 has parallax 0.85″ from opposite sides of Earth's orbit (2 AU apart).

• Use the right triangle formed by bisecting the parallax angle.
• Distance x = 1/tan(0.425″) ≈ 485,000 AU.

Don't confuse: When the parallax angle is very small, the triangle is nearly isosceles, and you can often use a right triangle approximation by bisecting the angle.

🧭 Overview

🧠 One-sentence thesis

The Law of Cosines generalizes the Pythagorean theorem to all triangles and enables us to solve triangles when we know two sides and the included angle (SAS) or all three sides (SSS), situations where the Law of Sines is not helpful.

📌 Key points (3–5)

• When to use the Law of Cosines: when you know two sides and the included angle (SAS), or when you know all three sides (SSS)—cases where the Law of Sines fails.
• What the Law of Cosines says: it relates all three sides and one angle; when the angle is 90°, it reduces to the Pythagorean theorem.
• How side c changes with angle C: if angle C is acute, c² is less than a² + b²; if C is obtuse, c² is greater than a² + b²; if C is exactly 90°, c² equals a² + b².
• Common confusion—order of operations: the Law of Cosines has three terms; the coefficient of the cosine term must be multiplied by the cosine value before subtracting, not subtracted from the previous sum first.
• Ambiguous case alternative: the Law of Cosines can solve the ambiguous case (SSA) by finding the third side first using the quadratic formula; the number of positive solutions tells you how many triangles exist.

🔍 When the Law of Sines fails

🔍 Two sides and the included angle

Included angle: the angle between two known sides.

• The excerpt contrasts two scenarios:
  • Law of Sines works: you know two sides and the angle opposite one of them (e.g., sides a = 5 and c = 7, and angle A = 28°).
  • Law of Sines does not work: you know two sides and the included angle (e.g., sides a = 5 and c = 7, and angle B = 115° between them).
• In the second case, you cannot set up a proportion with the Law of Sines because you don't know the angle opposite either known side.
• Example: To find the distance from Avery to Clio, you know two sides (a = 34 and c = 48) and the included angle (B = 125°). The Law of Sines cannot help here.

🔍 Why we need a generalization

• The Pythagorean theorem (c² = a² + b²) only works when angle C is exactly 90°.
• The Law of Cosines extends this relationship to all triangles, no matter the size of angle C.

📐 The Law of Cosines formula

📐 Three versions of the law

Law of Cosines: If the angles of a triangle are A, B, and C, and the opposite sides are respectively a, b, and c, then:

• a² = b² + c² − 2bc cos(A)
• b² = a² + c² − 2ac cos(B)
• c² = a² + b² − 2ab cos(C)

• Each version solves for the square of one side in terms of the other two sides and the angle opposite the unknown side.
• When the angle is 90°, cos(90°) = 0, so the "−2ab cos(C)" term vanishes and the equation becomes the Pythagorean theorem.

📐 How the formula changes with angle size

The excerpt shows three cases for a triangle with fixed sides a and b but varying angle C:

Angle C	Relationship	Explanation
C = 90°	c² = a² + b²	The Pythagorean theorem; the cosine term is zero
C is acute	c² < a² + b²	Side c is shorter; the cosine of an acute angle is positive, so subtracting 2ab cos(C) makes c² smaller
C is obtuse	c² > a² + b²	Side c is longer; the cosine of an obtuse angle is negative, so subtracting a negative (i.e., adding) makes c² larger

• Don't confuse: the formula always has a minus sign before 2ab cos(C); the sign of cos(C) itself determines whether you effectively subtract or add.

🧮 Finding a side with the Law of Cosines

🧮 Step-by-step procedure

1. Identify the two known sides and the included angle.
2. Choose the version of the Law of Cosines that uses the known angle.
3. Substitute the known values.
4. Simplify the right side (following order of operations carefully).
5. Take the positive square root to find the unknown side.

🧮 Worked example: distance from Avery to Clio

• Given: a = 34, c = 48, and B = 125°. Find b.
• Use the version: b² = a² + c² − 2ac cos(B).
• Substitute: b² = 34² + 48² − 2(34)(48) cos(125°).
• Simplify: b² = 1156 + 2304 − 3264 cos(125°) = 3460 − 3264(−0.573567...) = 5332.153.
• Take the square root: b ≈ 73.02 miles.

⚠️ Order of operations caution

The excerpt warns:

• The right side has three terms: 34², 48², and −2(34)(48) cos(125°).
• First compute 34² = 1156 and 48² = 2304, giving 3460.
• Then compute 2(34)(48) = 3264, which is the coefficient of cos(125°).
• Multiply 3264 by cos(125°) ≈ −0.5736, giving −1872.15.
• Finally subtract: 3460 − (−1872.15) = 3460 + 1872.15 = 5332.15.
• Wrong approach: subtracting 3264 from 3460 first would give 196, which is incorrect.
• Tip: On a graphing calculator, you can enter the entire right side exactly as written.

🔢 Finding an angle with the Law of Cosines

🔢 When to use this approach

• Use the Law of Cosines to find an angle when you know all three sides of the triangle (SSS case).

🔢 Algebraic steps

1. Choose the version of the Law of Cosines that uses the unknown angle.
2. Substitute the three known side lengths.
3. Simplify both sides.
4. Isolate the cosine term (move other terms to the left side).
5. Solve for cos(angle).
6. Apply the inverse cosine function to find the angle.

🔢 Worked example: finding angle C

• Given: a = 6, b = 7, c = 11. Find angle C.
• Use: c² = a² + b² − 2ab cos(C).
• Substitute: 11² = 6² + 7² − 2(6)(7) cos(C).
• Simplify: 121 = 36 + 49 − 84 cos(C) → 121 = 85 − 84 cos(C).
• Isolate: 121 − 85 = −84 cos(C) → 36 = −84 cos(C).
• Solve: cos(C) = 36 / (−84) = −3/7.
• Find C: C = cos⁻¹(−3/7) ≈ 115.4°.

🔢 Finding a second angle

After finding one angle, you can find a second angle using either:

• Law of Sines (fewer calculations, but uses rounded values from previous steps).
• Law of Cosines (more calculations, but uses only original given values, so more accurate).

The excerpt recommends the Law of Cosines for accuracy, because:

• Each rounding introduces inaccuracy.
• Using calculated (rounded) values compounds the error.
• Using only given values minimizes accumulated error.

Another advantage of the Law of Cosines:

• There is only one angle between 0° and 180° with a given cosine, so no need to check for a second possible angle (unlike the Law of Sines, where two angles can have the same sine).

🔀 Using the Law of Cosines for the ambiguous case

🔀 What is the ambiguous case

• You know two sides (a and b) and an acute angle α opposite one of them (SSA).
• Depending on the relative sizes of a, b, and α, there may be:
  • No solution: a is too short to reach the opposite side.
  • One solution: a is exactly the right length (forms a right triangle), or a is longer than b.
  • Two solutions: a is long enough to reach the opposite side in two different places.

The excerpt lists four scenarios:

Condition	Number of solutions	Description
a < b sin(α)	0	Side a is too short to make a triangle
a = b sin(α)	1	Side a makes exactly a right triangle
b sin(α) < a < b	2	Side a can form two different triangles
a > b	1	Side a is long enough for one triangle

• If α is obtuse: one solution if a > b, no solution if a ≤ b.

🔀 Law of Cosines approach: find the third side first

Instead of using the Law of Sines (which requires checking both possible angles), you can:

1. Use the Law of Cosines to find the third side.
2. This leads to a quadratic equation.
3. Apply the quadratic formula.
4. The number of positive solutions tells you how many triangles exist.

Quadratic formula: The solutions of ax² + bx + c = 0 (a ≠ 0) are x = (−b ± √(b² − 4ac)) / (2a).

Discriminant: b² − 4ac. It determines the number of solutions.

• One positive solution → one triangle.
• Two positive solutions → two triangles.
• No positive solutions → no triangle.

🔀 Worked example: two triangles

• Given: B = 14.4°, a = 8, b = 3. Solve the triangle.
• Use: b² = a² + c² − 2ac cos(B).
• Substitute: 3² = 8² + c² − 2(8)c cos(14.4°).
• Simplify: 9 = 64 + c² − 16c(0.9686) → 9 = 64 + c² − 15.497c.
• Standard form: c² − 15.497c + 55 = 0.
• Quadratic formula: c = (15.497 ± √((−15.497)² − 4(1)(55))) / 2 = (15.497 ± 4.490) / 2.
• Two solutions: c ≈ 5.503 or c ≈ 9.994.
• Because there are two positive solutions, there are two triangles.
• For each value of c, use the Law of Cosines again to find angle C, then find angle A by subtraction (A = 180° − B − C).
• Triangle 1: c ≈ 5.503, C ≈ 27.1°, A ≈ 138.5°.
• Triangle 2: c ≈ 9.994, C ≈ 124.1°, A ≈ 41.5°.

🔀 Why the Law of Cosines avoids ambiguity

• When finding an angle using the Law of Cosines, there is only one angle between 0° and 180° with a given cosine.
• You don't need to check for a second possible angle, as you would with the Law of Sines.

🧭 Navigation applications

🧭 Real-world context

• Aircraft pilots and ship captains need to understand trigonometry-based navigation, even with GPS.
• Navigation problems often involve headings (directions measured as angles from north) and distances.

🧭 Worked example: sailing club rescue

• The sailing club sails 18 miles on a heading 15° east of north, then 12 miles on a heading 35° east of north, then calls for help.
• Find the distance x and heading θ for the speed boat to travel from the marina.
• In triangle ABC, angle B = 180° − (35° − 15°) = 160°.
• Known: a = 12, c = 18, B = 160°.
• Use Law of Cosines: b² = a² + c² − 2ac cos(B) = 12² + 18² − 2(12)(18) cos(160°) = 873.95 → b ≈ 29.56 miles.
• Use Law of Cosines again to find angle A: cos(A) = (12² − 29.56² − 18²) / (−2(29.56)(18)) ≈ 0.9903 → A ≈ 8°.
• Heading: θ = 8° + 15° = 23° east of north.
• Answer: The speed boat should travel 29.56 miles on a heading 23° east of north.

🗂️ Deciding which law to use

🗂️ Decision table

The excerpt provides a clear guide:

What you know	Abbreviation	Which law to use	What to find first
One side and two angles	SAA	Law of Sines	Another side
Two sides and the angle opposite one of them	SSA (ambiguous case)	Law of Sines (to find an angle) or Law of Cosines (to find a side)	Check both approaches
Two sides and the included angle	SAS	Law of Cosines	The third side
Three sides	SSS	Law of Cosines	An angle

• Right triangles: Don't need either law; use the basic trigonometric ratios (sine, cosine, tangent).
• Oblique triangles: Use the table above.

🗂️ Example: which law for which part?

• Given: angle A = 40°, side b = 6, side c = 10. Find angle B and side a.
• This is the ambiguous case (SSA): two sides and the angle opposite one of them.
• To find angle B: use the Law of Sines. sin(B)/6 = sin(40°)/10 → sin(B) ≈ 0.3857. Two angles have this sine; check each.
• To find side a: use the Law of Cosines. 10² = 6² + a² − 2(6)a cos(40°) → a² − (12 cos 40°)a − 64 = 0. Solve the quadratic equation.
• The excerpt notes: because a > b, angle B must be acute, so only one triangle exists (even though the sine equation gives two angles, only one is valid).

🗂️ Another example: SAS case

• Given: angle C = 115°, side a = 71, side b = 53. What can you find, and which law?
• This is the SAS case: two sides and the included angle.
• First find side c using the Law of Cosines: c² = a² + b² − 2ab cos(C).

📚 Summary concepts

📚 Key vocabulary

• Quadratic equation: an equation of the form ax² + bx + c = 0.
• Quadratic formula: x = (−b ± √(b² − 4ac)) / (2a).
• Discriminant: the expression b² − 4ac under the square root; it determines the number of solutions.

📚 When the Law of Sines is not helpful

• The Law of Sines requires knowing an angle and its opposite side, plus one other side or angle.
• When you know two sides and the included angle (SAS), you don't have an angle-opposite-side pair to start with.
• The Law of Cosines fills this gap.

📚 Relationship to the Pythagorean theorem

• The Pythagorean theorem is a special case of the Law of Cosines.
• When angle C = 90°, cos(90°) = 0, so the term −2ab cos(C) disappears, leaving c² = a² + b².

📚 Accuracy tip

• Whenever possible, use given values instead of calculated (rounded) values in subsequent steps.
• Each rounding introduces error; using only original values minimizes accumulated inaccuracy.
• This is why the Law of Cosines is preferred over the Law of Sines when finding a second angle after you've already calculated one.

🧭 Overview

🧠 One-sentence thesis

Angles can describe rotation beyond triangles, and by placing any angle in standard position we can define its trigonometric ratios using coordinates, with reference angles allowing us to relate all angles back to familiar first-quadrant values.

📌 Key points (3–5)

• Angles as rotation: Angles measure rotation in degrees; one complete rotation is 360°, and positive angles rotate counter-clockwise while negative angles rotate clockwise.
• Standard position definition: Any angle's sine, cosine, and tangent are defined by placing the angle at the origin with initial side on the positive x-axis, then using coordinates (x, y) and distance r from a point on the terminal side.
• Reference angles: Every angle has a reference angle (the acute angle to the x-axis), and the trig ratios of any angle equal those of its reference angle except for sign, which depends on the quadrant.
• Common confusion: Coterminal angles (differing by multiples of 360°) have identical trig ratios because they share the same terminal side, even though their degree measures differ.
• Special angles: The angles 30°, 45°, and 60° (and their counterparts in other quadrants) allow exact radical expressions for trig ratios.

🔄 Rotation and standard position

🔄 Measuring rotation with angles

• Angles measure how far something rotates, not just parts of triangles.
• One complete rotation = 360°.
• Example: A clock's minute hand rotates 360° in one hour, so in 1.5 hours it rotates 1.5 × 360° = 540°, and in 40 minutes (two-thirds of an hour) it rotates (2/3) × 360° = 240°.

📐 Standard position

Standard position: The vertex of the angle is at the origin, and the initial side lies on the positive x-axis.

• Positive angles: terminal side rotates counter-clockwise.
• Negative angles: terminal side rotates clockwise.
• This convention allows us to compare and analyze angles consistently.
• Example: An angle of 320° in standard position has its terminal side in the fourth quadrant (between 270° and 360°).

🗺️ Quadrants and degree ranges

Quadrant	Degree range	Terminal side location
First	0° < θ < 90°	Upper right
Second	90° < θ < 180°	Upper left
Third	180° < θ < 270°	Lower left
Fourth	270° < θ < 360°	Lower right

• 180° is one-half a complete revolution; 270° is three-quarters.
• Example: An angle of 200° lies in the third quadrant because 180° < 200° < 270°.

📏 Trigonometric ratios for all angles

📏 Defining trig ratios with coordinates

The Trigonometric Ratios: If θ is an angle in standard position, and (x, y) is a point on its terminal side, with r = √(x² + y²), then
sin(θ) = y/r
cos(θ) = x/r
tan(θ) = y/x

• Why this works: You can choose any point P on the terminal side; the ratios will be the same because of similar triangles.
• r is always positive (it's a distance), but x and y can be positive, negative, or zero depending on the quadrant.
• Example: For a point P at (−√11, −5) with r = 6 in the third quadrant, sin(θ) = −5/6, cos(θ) = −√11/6, tan(θ) = 5/√11.

➕➖ Signs of trig ratios by quadrant

Quadrant	x	y	sin	cos	tan
I	+	+	+	+	+
II	−	+	+	−	−
III	−	−	−	−	+
IV	+	−	−	+	−

• Quadrant I: All positive.
• Quadrant II: Sine positive (y > 0, r > 0).
• Quadrant III: Tangent positive (both x and y negative, so y/x is positive).
• Quadrant IV: Cosine positive (x > 0, r > 0).
• Don't confuse: The sign depends on the quadrant, not the size of the angle.

🔗 Reference angles

🔗 What is a reference angle?

Reference angle: The positive acute angle formed between the terminal side of θ and the x-axis.

• How to construct a reference triangle:
  1. Choose a point P on the terminal side.
  2. Draw a perpendicular line from P to the x-axis.
• The reference triangle always lies between the terminal side and the x-axis.
• Example: The reference angle for 130° is 180° − 130° = 50°, because 130° is in the second quadrant.

🧮 Finding reference angles by quadrant

Quadrant	Formula for reference angle θ̃
First	θ̃ = θ
Second	θ̃ = 180° − θ
Third	θ̃ = θ − 180°
Fourth	θ̃ = 360° − θ

• Example: For θ = 200° (third quadrant), θ̃ = 200° − 180° = 20°.
• Example: For θ = 285° (fourth quadrant), θ̃ = 360° − 285° = 75°.

🔁 Using reference angles to find trig ratios

• Key principle: The trig ratios of any angle equal the ratios of its reference angle, except for sign (determined by the quadrant).
• Example: sin(200°) = −sin(20°) because 200° is in the third quadrant (sine negative) with reference angle 20°.
• Example: cos(285°) = cos(75°) because 285° is in the fourth quadrant (cosine positive) with reference angle 75°.

🎯 Finding angles from reference angles

Given a reference angle θ̃, the angles in each quadrant are:

Quadrant	Formula
I	θ = θ̃
II	θ = 180° − θ̃
III	θ = 180° + θ̃
IV	θ = 360° − θ̃

• These four angles together make a "bow-tie" shape.
• Example: If the reference angle is 35°, the four angles are 35°, 145°, 215°, and 325°.
• Important: There are always two angles between 0° and 360° with a given trig ratio (except for quadrantal angles like 0°, 90°, 180°, 270°).

🔍 Solving for angles with given trig ratios

Example: Find two angles between 0° and 360° whose cosine is 5/8.

• First, find the first-quadrant angle: cos⁻¹(5/8) ≈ 51.3°.
• Cosine is also positive in the fourth quadrant, so the second angle is 360° − 51.3° = 308.7°.

Example: Find two angles whose cosine is −5/8.

• The reference angle is still 51.3°.
• Cosine is negative in the second and third quadrants.
• Second quadrant: 180° − 51.3° = 128.7°.
• Third quadrant: 180° + 51.3° = 231.3°.

Don't confuse: Use the sign of the trig ratio to determine which quadrants to check.

⭐ Special angles

⭐ The special angles in all quadrants

• Special angles: 30°, 45°, and 60° (and their counterparts in other quadrants).
• These angles allow exact values using radicals, based on 30-60-90 and 45-45-90 triangles.
• The special angles in all four quadrants are: 30°, 45°, 60°, 120°, 135°, 150°, 210°, 225°, 240°, 300°, 315°, 330°.

🧮 Exact values for special angles

Example: Find exact values for sin(210°), cos(210°), tan(210°).

• 210° is in the third quadrant with reference angle 30°.
• In the third quadrant, sine and cosine are negative.
• sin(210°) = −1/2, cos(210°) = −√3/2, tan(210°) = 1/√3.

Alternative method: Use the reference triangle with r = 2. For a 30-60-90 triangle, the sides are in ratio 1 : √3 : 2, so P has coordinates (−√3, −1), giving the same results.

Example: For 300° (fourth quadrant, reference angle 60°):

• sin(300°) = −√3/2 (sine negative in fourth quadrant).
• cos(300°) = 1/2 (cosine positive in fourth quadrant).
• tan(300°) = −√3 (tangent negative in fourth quadrant).

🔄 Coterminal angles

🔄 What are coterminal angles?

Coterminal angles: Angles that differ by a multiple of 360° and thus have the same terminal side.

• Example: 70°, 430° (= 70° + 360°), and 790° (= 70° + 2 × 360°) are all coterminal.
• Key property: Coterminal angles have equal trigonometric ratios because they share the same standard position.
• Example: cos(790°) = cos(70°) = 0.3420.

🔢 Finding coterminal angles

• To find a coterminal angle between 0° and 360°: Add or subtract multiples of 360° until the result is in that range.
• Example: For 520°, subtract 360° to get 520° − 360° = 160°.
• Example: For −60°, add 360° to get −60° + 360° = 300°.

➕➖ Positive and negative angles

• Positive angles: counter-clockwise rotation.
• Negative angles: clockwise rotation.
• Example: −60° (clockwise) is coterminal with 300° (counter-clockwise).
• Don't confuse: The direction matters when describing rotation, but coterminal angles have the same trig ratios regardless of direction.

🧭 Overview

🧠 One-sentence thesis

Sine, cosine, and tangent functions produce periodic graphs that model cyclical phenomena, and their shapes are controlled by parameters that determine amplitude, period, and midline.

📌 Key points (3–5)

• Periodic behavior: Trigonometric functions repeat their values in regular intervals, making them ideal for modeling cyclical phenomena like Ferris wheel rides, tides, and planetary motion.
• Sine and cosine graphs: Both have the same wave shape with period 360°, amplitude 1, and midline y = 0, but differ in where they start and where their maximum/minimum values occur.
• Tangent graph: Unlike sine and cosine, tangent has period 180°, no amplitude, vertical asymptotes at odd multiples of 90°, and increases continuously on each interval of its domain.
• Parameters control shape: The values A, B, and k in equations like y = A sin(Bθ) + k determine amplitude (|A|), period (360°/|B|), and midline (y = k).
• Common confusion: Don't confuse cos(θ) as "cos times θ"—it represents a single output value from the cosine function, not a product.

🎡 Modeling with periodic functions

🎡 The Ferris wheel example

The excerpt introduces periodic functions through a concrete scenario: riding a Ferris wheel with radius 100 feet.

• Setup: Place the origin at the wheel's center; your position makes angle θ with the horizontal.
• Height relationship: Your height h above ground relates to your y-coordinate by h = y + 100 (since the center is 100 feet up).
• The sine connection: From the circle geometry, sin(θ) = y/100, so y = 100 sin(θ).
• Complete model: h = 100 sin(θ) + 100 describes height as a function of angle.

Example: At θ = 30°, y = 100 sin(30°) = 100(1/2) = 50 feet, so h = 150 feet above ground.

🔄 What makes a function periodic

Periodic function: A function that repeats the same pattern at regular intervals.

• The Ferris wheel height repeats every complete rotation (360°).
• Period: The smallest interval on which the graph repeats.
• This cyclical behavior appears in tides, plant growth cycles, radio waves, and planetary motion.

📈 The sine and cosine functions

📈 Graphing sine: f(θ) = sin(θ)

The sine function gives the y-coordinate of a point traveling around a unit circle (radius r = 1).

Behavior by quadrant:

• 0° to 90°: sin(θ) increases from 0 to 1
• 90° to 180°: sin(θ) decreases from 1 to 0
• 180° to 270°: sin(θ) decreases from 0 to −1
• 270° to 360°: sin(θ) increases from −1 to 0

Key properties:

• Period: 360°
• Maximum value: 1; minimum value: −1
• Midline: y = 0 (the horizontal line the graph oscillates around)
• Amplitude: 1 (distance from midline to maximum or minimum)

📉 Graphing cosine: f(θ) = cos(θ)

The cosine function gives the x-coordinate of a point traveling around a unit circle.

Behavior by quadrant:

• 0° to 90°: cos(θ) decreases from 1 to 0
• 90° to 180°: cos(θ) decreases from 0 to −1
• 180° to 270°: cos(θ) increases from −1 to 0
• 270° to 360°: cos(θ) increases from 0 to 1

Key properties: Same period (360°), amplitude (1), and midline (y = 0) as sine.

How cosine differs from sine: The cosine graph starts (at θ = 0°) at its high point (maximum value 1), while the sine graph starts at its midline (value 0).

🚴 The bicycle example

The excerpt uses a bicycle pedal to illustrate cosine: when the crank is 18 cm long and makes angle θ with horizontal, the foot's x-coordinate is x = 18 cos(θ).

• The horizontal displacement from a vertical reference line (KOPS line at x = 18) is d = 18 − 18 cos(θ).
• This function has period 360°, amplitude 18 cm, and midline d = 18.
• The cosine naturally describes horizontal position, just as sine describes vertical position.

📐 The tangent function

📐 How tangent differs

The tangent function is defined by tan(θ) = y/x, where (x, y) is a point on the terminal side of angle θ.

Unique characteristics:

• Period: 180° (not 360° like sine and cosine)
• No amplitude: The function grows without bound; it does not oscillate between fixed maximum and minimum values
• Undefined values: tan(θ) is undefined at 90°, 270°, and all their coterminal angles (odd multiples of 90°)
• Vertical asymptotes: The graph has breaks at θ = 90°, θ = 270°, etc.
• Always increasing: On each interval where it is defined, tangent increases continuously

📐 Behavior by quadrant

• First quadrant (0° to 90°): tan(θ) increases from 0 toward +∞ as θ approaches 90°
• Second quadrant (90° to 180°): tan(θ) is negative; it increases from −∞ toward 0
• Third and fourth quadrants: The pattern repeats because tan(θ + 180°) = tan(θ)

Example: tan(130°) = −tan(50°) because 130° is in the second quadrant with reference angle 50°.

Don't confuse: Tangent does not have an amplitude because it does not stay within fixed bounds.

🎛️ Parameters that control graphs

🎛️ Amplitude: the A parameter

In y = A sin(θ) or y = A cos(θ), the amplitude is |A|.

• What it does: Stretches or compresses the graph vertically
• The graph oscillates between −|A| and +|A|

Example: y = 3 sin(θ) has amplitude 3; it oscillates between −3 and +3, while y = sin(θ) oscillates between −1 and +1.

🎛️ Period: the B parameter

In y = sin(Bθ) or y = cos(Bθ), the period is 360°/|B|.

• What it does: Controls how quickly the function completes one cycle
• Larger B → shorter period → more cycles in the same interval

Example: y = sin(3θ) has period 360°/3 = 120°; it completes three full cycles from 0° to 360°, while y = sin(θ) completes only one.

🎛️ Midline: the k parameter

In y = k + sin(θ) or y = k + cos(θ), the midline is the horizontal line y = k.

• What it does: Shifts the entire graph up or down
• The graph oscillates around y = k instead of y = 0

Example: y = 3 + sin(θ) has midline y = 3; the graph oscillates between 2 and 4, shifted up 3 units from the standard sine graph.

🎛️ Combined parameters

General form: y = k + A sin(Bθ) or y = k + A cos(Bθ)

• Amplitude: |A|
• Period: 360°/|B|
• Midline: y = k

Example: For y = −3 + 4 sin(3θ):

• Amplitude = 4
• Period = 360°/3 = 120°
• Midline: y = −3
• The graph oscillates between −7 and 1

🔤 Function notation review

🔤 Input and output variables

The notation y = f(x) means "y is a function of x," where x is the input and y is the output.

• For trigonometric functions, θ is typically the input (the angle)
• The output is the value of sin(θ), cos(θ), or tan(θ)

Variable names can change: The function y = f(x) = √(9 − x²) is the same as w = f(t) = √(9 − t²); only the labels differ, not the relationship.

🔤 Trigonometric function notation

When working with trig functions, multiple variables are involved: x, y, r, and θ.

• If r is fixed (e.g., r = 1 for the unit circle), then x and y are both functions of θ
• We write x = f(θ) = cos(θ) and y = g(θ) = sin(θ)
• Graph axes: The horizontal axis shows the input (θ values); the vertical axis shows the output (function values)

🔤 Common notation mistake

Don't confuse: cos(θ) is not a product "cos times θ."

• "cos" by itself has no meaning; it is not a number
• cos(θ) represents a single number: the output of the cosine function when the input is θ
• Similarly for sin(θ) and tan(θ)

Example: If you write d = F(φ) = sin(φ), the graph should have φ on the horizontal axis and d on the vertical axis.

🔍 Comparing the three functions

Feature	Sine	Cosine	Tangent
Period	360°	360°	180°
Amplitude	1	1	None (unbounded)
Midline	y = 0	y = 0	None
Value at θ = 0°	0	1	0
Undefined at	Never	Never	90°, 270°, ...
Shape	Wave starting at midline	Wave starting at maximum	Increasing curves with asymptotes

Key distinction: Sine and cosine have similar wave shapes but are shifted horizontally; tangent has a completely different shape with vertical asymptotes and no upper or lower bounds.

🧭 Overview

🧠 One-sentence thesis

Trigonometric functions extend beyond angle measurement to solve equations, locate points in coordinate systems, and model real-world periodic phenomena such as tides, daylight hours, and orbital motion.

📌 Key points (3–5)

• Solving trig equations: A calculator gives only one solution; reference angles and coterminal angles help find all solutions between 0° and 360°.
• Unit circle coordinates: On a unit circle, any point determined by angle θ has coordinates (cos θ, sin θ), which scales to (r cos θ, r sin θ) on a circle of radius r.
• Bearings and navigation: Bearings measure angles clockwise from north; converting to standard position requires θ = –bearing + 90°.
• Common confusion: Calculators return one angle; you must determine which quadrant(s) contain solutions and use reference angles to find all answers.
• Periodic functions: Any function that repeats at fixed intervals is periodic; sinusoidal functions (shaped like sine or cosine) model many natural cycles.

🔍 Solving Trigonometric Equations

🔍 Why reference angles matter

• Between 0° and 180°, there are always two angles with the same sine value (except at 0° and 180°).
• Example: sin θ = 0.4226 has solutions 25° and 155° (supplementary angles share a reference angle).
• For negative sine values, solutions lie in the third and fourth quadrants.

🧮 Calculator limitations

Caution: When you ask your calculator to find an angle with a given sine, cosine, or tangent, it will give you only one of the many possible answers.

• The calculator may return a negative angle or an angle outside 0°–360°.
• You must find coterminal angles (add or subtract 360°) and use reference angles to locate all solutions in the desired range.
• Example: sin⁻¹(–0.6428) = –40°; add 360° to get 320° (fourth quadrant), then find the third-quadrant solution 220° using the reference angle 40°.

📐 Finding all solutions

Steps:

1. Use the inverse trig function on your calculator to get one solution.
2. If the angle is outside 0°–360°, add or subtract 360° to find a coterminal angle.
3. Identify the reference angle.
4. Determine which other quadrant(s) have the same trig ratio (with appropriate sign).
5. Construct the second solution using the reference angle.

Example scenario: To find when a Ferris wheel rider is at height 136 feet (given h = 100 + 100 sin θ), solve sin θ = 0.36. One solution is 21.1°; the second is 180° – 21.1° = 158.9°.

🔵 The Unit Circle

🔵 What makes it special

Unit circle: A circle of radius 1 centered at the origin.

• On a unit circle, the hypotenuse of the reference triangle is r = 1.
• The definitions simplify: cos θ = x/1 = x and sin θ = y/1 = y.
• Therefore, any point P on the unit circle determined by angle θ has coordinates (cos θ, sin θ).

📏 Scaling to any radius

• On a circle of radius r, the reference triangle is scaled by factor r.
• The coordinates become (r cos θ, r sin θ).
• This formula locates any point at distance r from the origin in the direction specified by θ.

Example: A point 6 cm from the origin at angle 292° has coordinates:

• x = 6 cos(292°) ≈ 2.25
• y = 6 sin(292°) ≈ –5.56

🗺️ Finding exact coordinates

• For special angles (30°, 45°, 60°, etc.), use known trig ratios.
• Example: At 315° (fourth quadrant, reference angle 45°), cos(315°) = 1/√2 and sin(315°) = –1/√2, so the point on the unit circle is (1/√2, –1/√2).

🧭 Bearings and Navigation

🧭 What bearings measure

Bearings: Navigational directions given as angles measured clockwise from north.

• A bearing of 110° points 110° clockwise from north.
• To convert to standard position (counterclockwise from east): θ = –bearing + 90° (or add 360° if needed).

✈️ Using bearings to find position

Scenario: A plane flies 60 miles at bearing 245°.

1. Convert: θ = –245° + 90° = –155° (or 205°).
2. The southward distance is the y-coordinate: y = 60 sin(205°) ≈ –25.4 miles.
3. The plane is about 25.4 miles south of the airport.

Key insight: Bearings simplify real-world navigation but require conversion to standard position for calculation.

📐 Angle of Inclination

📐 Connecting slope and angle

Angle of inclination: The angle α measured counterclockwise from the positive x-axis to a line, where 0° ≤ α ≤ 180°.

• The slope m of a line equals the tangent of its angle of inclination: tan α = m.
• Example: A line with slope 3/4 has angle of inclination tan⁻¹(3/4) ≈ 36.9°.

📈 Understanding the tangent graph

• As α increases from 0° toward 90°, the slope increases toward infinity (steeper and steeper upward).
• As α decreases from 180° toward 90°, the slope decreases toward negative infinity (steeper and steeper downward).
• The tangent function is undefined at 90° (vertical lines have no slope).
• This behavior explains why tan θ has vertical asymptotes at odd multiples of 90°.

🌊 Periodic and Sinusoidal Functions

🌊 What makes a function periodic

Periodic function: A function y = f(x) is periodic if there is a smallest value p such that f(x + p) = f(x) for all x. The constant p is the period.

• The function repeats its pattern every p units.
• Not all periodic functions are sinusoidal (sine/cosine-shaped).
• Example: A sawtooth wave or a square wave can be periodic but not sinusoidal.

🌅 Real-world sinusoidal models

Characteristics of sinusoidal graphs:

• Period: The interval over which the pattern repeats.
• Midline: The horizontal line halfway between maximum and minimum values.
• Amplitude: The vertical distance from the midline to the maximum (or minimum).

Common applications:

• Daylight hours over a year (period ≈ 12 months).
• Tides (period ≈ 12 hours).
• Ferris wheel height (period = one rotation time).
• Temperature cycles (period = 1 year or 1 day).

🎡 Recognizing sinusoidal vs. other patterns

• Sinusoidal: Smooth, wave-like; changes slowly near extremes, rapidly near the midline.
• Linear periodic: Straight segments (e.g., a triangle wave); constant rate of change within each segment.
• Don't confuse: A graph that oscillates is not necessarily sinusoidal—check whether the rate of change is constant or varies smoothly.

Example: A minute hand's height changes sinusoidally because it moves in a circle; the height decreases rapidly near the quarter-hours and slowly near the hour and half-hour.

📊 Analyzing Periodic Graphs

📊 Reading period, amplitude, midline

From a graph:

1. Period: Measure the horizontal distance for one complete cycle.
2. Midline: Average of maximum and minimum values.
3. Amplitude: Distance from midline to maximum (or minimum).

Example (daylight in Glasgow):

• Period: 12 months.
• Midline: y ≈ 12.25 hours.
• Amplitude: ≈ 5.25 hours.

🩺 Non-sinusoidal periodic examples

• Medication dosing: Sharp jumps when doses are administered, exponential decay between doses; periodic but not sinusoidal.
• Bicycle light: Traces a cycloid (period = wheel circumference / speed); periodic but not a pure sine wave.

Key point: Identify the repeating interval and the pattern within each cycle; not every periodic function fits y = k + A sin(Bθ).

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Summary: This section shows how trigonometric functions solve equations (using reference angles and coterminal angles), locate points (via the unit circle and coordinate formulas), support navigation (through bearings), connect geometry and algebra (angle of inclination), and model cyclical real-world phenomena (periodic and sinusoidal functions).

🧭 Overview

🧠 One-sentence thesis

Trigonometric ratios behave like algebraic variables, so we can simplify, evaluate, multiply, and factor expressions containing them using the same techniques we use for ordinary algebra.

📌 Key points (3–5)

• Trig ratios are numbers, not products: sin(X) is a single output value from the sine function, not "sin times X."
• Treat each trig ratio as one variable: cos(θ) and sin(θ) are separate variables; combine like terms the same way you would with x and y.
• Order of operations matters: sin(3X + 45°) is not equal to sin(3X) + sin(45°); evaluate inside parentheses first.
• Common confusion—powers vs. angle operations: cos²(θ) means (cos(θ))² (square the output), not cos(θ²) (square the angle first).
• All standard algebra techniques apply: distributive law, FOIL, factoring quadratics, and combining like terms work exactly as they do with ordinary variables.

🔢 Evaluating trigonometric expressions

🔢 What evaluation means

Evaluate: substitute specific angle values into an expression and compute the result using known trig ratios.

• Replace each variable with its given angle measure.
• Compute each trig ratio (e.g., sin(30°) = 1/2, tan(135°) = -1).
• Follow the order of operations: parentheses, exponents, multiplication/division, addition/subtraction.

Example: To evaluate 2 tan(Y) + 3 sin(X) for X = 30°, Y = 135°:

• Substitute: 2 tan(135°) + 3 sin(30°)
• Evaluate ratios: 2(-1) + 3(1/2)
• Simplify: -2 + 3/2 = -1/2

⚠️ Function notation, not multiplication

• sin(X) does not mean "sin times X."
• It means: apply the sine function to angle X and get a single number.
• This is why sin(90° + 45°) ≠ sin(90°) + sin(45°).

Don't confuse: sin(A + B) with sin(A) + sin(B)—the first evaluates the sum inside the function; the second adds two separate function outputs.

🧮 Simplifying trigonometric expressions

🧮 Combining like terms

Simplify: rewrite an expression in an equivalent but easier form by combining like terms and applying algebraic rules.

• Treat each distinct trig ratio as a separate variable.
• 3 tan(A) + 4 tan(A) = 7 tan(A) (like terms).
• tan(A) and cos(A) are not like terms (different functions).
• cos(t) and cos(w) are not like terms (different angles).

Example: Simplify 3 tan(A) + 4 tan(A) - 2 cos(A):

• Combine tan(A) terms: 7 tan(A) - 2 cos(A)
• Cannot combine further because tan and cos are different.

🔄 Equivalent expressions

Two expressions are equivalent if they yield the same value for every angle substitution.

Example: 2x² - 12x + 3x + 6 simplifies to 2x² - 9x + 6; both give the same result for any x.

🔺 Powers of trigonometric ratios

🔺 Notation for powers

cos²(θ) means (cos(θ))²—square the cosine value.

• Compute cos(θ) first, then square the result.
• Written as cos²(θ) to avoid excessive parentheses and to distinguish from cos(θ²).

⚠️ Power of output vs. power of input

Notation	Meaning	Example (θ = 30°)
cos²(θ)	(cos(θ))²	(√3/2)² = 3/4
cos(θ²)	cos of (θ squared)	cos(900°) = cos(180°) = -1

Don't confuse: cos²(30°) with cos(30²)—the first squares the output (3/4); the second squares the angle first (giving -1).

📐 Higher powers

• sin³(θ) = (sin(θ))³
• tan⁴(θ) = (tan(θ))⁴
• Always: the exponent applies to the output of the trig function.

✖️ Multiplying trigonometric expressions

✖️ Distributive law

Multiply trig expressions the same way you multiply algebraic expressions.

Example: cos(t)(3 cos(t) - 2)

• Distribute: cos(t) · 3 cos(t) - cos(t) · 2
• Result: 3 cos²(t) - 2 cos(t)

Compare with algebra: x(3x - 2) = 3x² - 2x

🔀 FOIL for binomials

Use First, Outside, Inside, Last to expand products like (4 sin(C) - 1)(3 sin(C) + 2).

• First: 4 sin(C) · 3 sin(C) = 12 sin²(C)
• Outside: 4 sin(C) · 2 = 8 sin(C)
• Inside: -1 · 3 sin(C) = -3 sin(C)
• Last: -1 · 2 = -2
• Combine: 12 sin²(C) + 5 sin(C) - 2

🧩 Factoring trigonometric expressions

🧩 Common factor extraction

Factor out the greatest common trig expression, just as you would factor out a common variable.

Example: 6 sin²(θ) - 9 sin(θ)

• Common factor: 3 sin(θ)
• Result: 3 sin(θ)(2 sin(θ) - 3)

Compare with algebra: 6w² - 9w = 3w(2w - 3)

🔢 Factoring quadratic trinomials

Factor expressions like tan²(α) - 3 tan(α) - 10 by treating tan(α) as a single variable.

• Find p and q such that pq = -10 and p + q = -3.
• Solution: p = -5, q = 2
• Result: (tan(α) - 5)(tan(α) + 2)

Example: 3 sin²(β) - 2 sin(β) - 1

• Factor as you would 3z² - 2z - 1
• Result: (3 sin(β) + 1)(sin(β) - 1)

🎯 Key principle

Every algebraic factoring technique (common factors, difference of squares, trinomial factoring) applies to trig expressions—just treat each trig ratio as a variable placeholder.

🧭 Overview

🧠 One-sentence thesis

Trigonometric equations can be solved by isolating the trig ratio, finding one solution (exactly or with a calculator), and then using reference angles to find all solutions between 0° and 360°.

📌 Key points (3–5)

• What solving means: finding all angle values that make the equation true, typically between 0° and 360°.
• Core method: isolate the trig ratio (sin, cos, or tan), solve for one angle, then use reference angles to find other solutions in different quadrants.
• How many solutions: most simple trig equations have two solutions in one cycle (0° to 360°), but quadratic trig equations may have up to four.
• Common confusion: don't stop at one solution—sine and cosine equations usually have a second solution in another quadrant; tangent solutions are always 180° apart.
• Special cases: equations may have no solution if the trig ratio falls outside its range (sine and cosine must be between -1 and 1).

🔧 Basic solving technique

🔧 The three-step process

The excerpt gives a systematic procedure:

1. Isolate the trigonometric ratio on one side of the equation.
2. Find one solution:
   • Give an exact answer if the ratio is a special value (30°, 45°, 60°, etc.).
   • Otherwise use inverse trig keys on a calculator.
3. Use reference angles to find a second solution if one exists.

Example: To solve 139 - 125 cos(θ) = 240:

• Isolate: cos(θ) = -0.808
• Calculator gives θ = 143.9° (second quadrant)
• Reference angle is 180° - 143.9° = 36.1°
• Third-quadrant solution: 180° + 36.1° = 216.1°

🎯 Why reference angles matter

• The same trig ratio value occurs in two quadrants (for sine and cosine).
• Reference angles help locate the second solution by identifying which quadrants share the same absolute value.
• Don't confuse: tangent equations are simpler—solutions differ by exactly 180°, so just add or subtract 180° from the first solution.

⚠️ When there's no solution

If a trig ratio is asked to equal a value outside its range, the equation has no solution.

• Sine and cosine only take values from -1 to 1.
• Example: sin(β) = 20 has no solution because sine can never be 20.

📊 Graphical approach

📊 Using graphs to solve

The excerpt shows that you can graph both sides of an equation separately and find intersections:

• Graph y = (left side expression) and y = (right side value).
• Intersection points give the solutions.
• Example: For the Ferris wheel problem, graphing h = 139 - 125 cos(2t/3) and the line h = 240 shows two intersection times.

🔍 What graphs reveal

• Trig equations have infinitely many solutions because trig functions repeat every 360°.
• The graph shows the pattern: intersections occur twice per cycle.
• We typically list only solutions in one cycle (0° to 360°) because all others are coterminal (differ by multiples of 360°).

🧮 Quadratic trigonometric equations

🧮 Extraction of roots method

When the equation involves a squared trig ratio, isolate the square first, then take square roots.

Example: 4 tan²(θ) + 3 = 15

• Isolate: tan²(θ) = 3
• Extract roots: tan(θ) = ±√3
• This gives four solutions because both +√3 and -√3 each occur in two quadrants.
• tan(θ) = √3 at 60° and 240°; tan(θ) = -√3 at 120° and 300°.

🔨 Factoring method

Some quadratic trig equations can be factored like algebraic quadratics.

Example: 2 cos²(θ) - cos(θ) - 1 = 0

• Factor: (2 cos(θ) + 1)(cos(θ) - 1) = 0
• Set each factor to zero: cos(θ) = -1/2 or cos(θ) = 1
• Solve each: θ = 120°, 240° (from the first) and θ = 0° (from the second).
• Total: three solutions.

Don't confuse: treat cos(θ) like a variable x when factoring, then solve the resulting simpler equations separately.

🌊 Real-world application: Snell's Law

🌊 Light refraction

The excerpt introduces Snell's Law for light bending between media:

sin(θ₁) / sin(θ₂) = v₁ / v₂

• θ₁ is the angle of incidence (incoming angle from the normal line).
• θ₂ is the angle of refraction (outgoing angle from the normal).
• The ratio v₁/v₂ is the index of refraction.

🔬 Solving refraction problems

Example: Light passes from water to glass with index 1.1 and 23° incidence angle.

• Set up: sin(23°) / sin(θ) = 1.1
• Isolate: sin(θ) = sin(23°) / 1.1 = 0.3552
• Solve: θ = sin⁻¹(0.3552) = 20.8°

Note: For Snell's Law, only use acute angles (the physical context determines one unique answer).

🎓 Common patterns and special cases

🎓 How many solutions to expect

Equation type	Typical number of solutions (0° to 360°)
sin(θ) = k or cos(θ) = k, where -1 < k < 1	Two solutions (one per relevant quadrant)
tan(θ) = k	Two solutions, always 180° apart
Quadrantal angles (sin or cos = ±1 or 0)	May have only one solution
sin²(θ) = k or cos²(θ) = k	Up to four solutions
Factored quadratic	Varies; solve each factor separately

⚙️ Calculator vs exact values

• Exact values: Use special angles (30°, 45°, 60°) when the trig ratio is √3/2, 1/2, √2/2, etc.
• Calculator: For other values, use inverse trig functions and round as instructed.
• Watch out: The calculator gives only one angle; you must find others using reference angles or the 180° rule for tangent.

🧭 Overview

🧠 One-sentence thesis

Trigonometric identities are equations involving trigonometric functions that hold true for all legitimate values of the variables, unlike conditional equations that are true only for specific values.

📌 Key points (3–5)

• What an identity is: an equation that is true for all legitimate values of the variables, not just some specific values.
• How identities differ from conditional equations: conditional equations are true only for certain values and false for others; identities are always true.
• Equivalent expressions: the two sides of an identity are equivalent expressions because they have the same value for all values of the variable.
• Common confusion: don't confuse identities (true for all values) with conditional equations (true only when you solve for specific values).
• Connection to simplification: when you simplify an algebraic expression, you are actually creating an identity.

🔍 Conditional equations vs identities

🔍 What a conditional equation is

• A conditional equation is true only for certain values of the variable and false for others.
• Example: the equation x squared plus 3x equals 10 is true only if x equals 2 or x equals 5.
• When you solve a conditional equation, you are finding the specific values of the variable that make the equation true.

✅ What an identity is

Identity: An identity is an equation that is true for all legitimate values of the variables.

• Unlike conditional equations, identities hold for every allowed value.
• The excerpt emphasizes "all legitimate values"—meaning any value for which the expressions are defined.
• Example: 3 times (x plus y) equals 3x plus 3y is an identity because it is true no matter what x and y are.

🔗 Equivalent expressions

🔗 Why identities create equivalent expressions

• In an identity, the expressions on either side of the equal sign are equivalent expressions.
• They have the same value for all values of the variable.
• This is the key property: both sides always produce the same result.

🧮 Connection to simplification

• When you simplify an algebraic expression, you are in fact creating an identity.
• Simplification operations (like combining like terms or applying the distributive law) produce equivalent expressions.
• Example: combining 3s plus 7s to get 10s creates the identity 3s plus 7s equals 10s, which is true for all values of s.

📋 Examples from the excerpt

📋 Algebraic identities

The excerpt provides several examples of identities:

Equation	Why it is an identity
3 times (x plus y) equals 3x plus 3y	Distributive law; true for all x and y
(x plus 1) squared equals x squared plus 2x plus 1	Expanding a binomial; true for all x
3s plus 7s equals 10s	Combining like terms; true for all s
5c times (c minus 2s) equals 5c squared minus 10cs	Distributive law; true for all c and s

📋 Non-identity example

• The equation 2t minus 1 equals 3 is not an identity.
• It is a conditional equation because it is true only for a specific value of t (when t equals 2).
• Don't confuse: just because an equation involves variables doesn't make it an identity; it must be true for all values.

🧭 Overview

🧠 One-sentence thesis

Radian measure connects angles directly to arclength on a circle, making it a more natural unit than degrees for mathematical work because it simplifies the relationship to s = rθ.

📌 Key points (3–5)

• Why radians exist: Degrees (360°) are arbitrary historical convention; radians arise naturally from the geometry of circles and the constant π.
• What a radian measures: One radian is the angle that spans an arc equal in length to the radius of the circle (about 57.3°).
• The arclength formula: When θ is in radians, arclength s = rθ (just radius times angle); no extra fractions needed.
• Unit circle special property: On a circle of radius 1, the arc length equals the radian measure of the angle (s = θ).
• Common confusion: Radian measure can be exact (multiples of π) or decimal approximations; both describe the same angle.

📜 Historical context and motivation

📜 Why 360 degrees?

• The Babylonians (around 600 BCE) likely originated the 360° circle.
• They used a base-60 number system because 60 has many factors (2, 3, 4, 5, 6), making division easier without decimal fractions.
• Six equilateral triangles fit around a point; each corner is 60°, so 6 × 60 = 360°.
• Babylonian astronomers calculated a year as 360 days, reinforcing the usefulness of 360°.
• Key insight: 360 is not fundamental to circles—it's historical accident.

📐 The natural unit: radians

• Developed by Roger Cotes (worked with Isaac Newton); he recognized radians as a natural unit.
• The term "radian" first appeared in print June 5, 1873, by James Thomson.
• Why radians matter: In calculus and advanced mathematics, radians make formulas simpler and more concise, especially for trigonometric functions.

🔄 Multiple ways to describe position on a circle

🔄 Four equivalent methods

The excerpt uses a Ferris wheel (radius 100 feet, 8-minute rotation) to show different ways to mark the same locations:

Method	At start	Quarter turn	Half turn	Three-quarters
Degrees θ	0°	90°	180°	270°
Percent p	0	25	50	75
Time t (min)	0	2	4	6
Arclength s (ft)	0	157	314	471

• Arclength method: Uses distance traveled along the circle's edge.
• This leads naturally to radian measure.

🎯 Key vocabulary

Arc: A portion of a circle.

Arclength: The length of an arc.

Central angle: An angle with vertex at the center of the circle.

Subtend (or span): A central angle subtends an arc when its sides meet the arc's endpoints.

Angle of displacement: A central angle representing distance traveled.

📏 Arclength calculation

📏 The basic principle

• Circumference of a circle: C = 2πr (about 6.28 times the radius).
• Arclength is proportional to the angle: the bigger the angle, the longer the arc.

Arclength formula (general): Arclength = (fraction of one revolution) · (2πr)

Example: An angle of 45° is 1/8 of a revolution, so s = (1/8)(2πr) = (π/4)r.

📏 Worked scenario

Example from excerpt: What arc does 120° span on a circle of radius 12 cm?

• 120° is 120/360 = 1/3 of a revolution.
• s = (1/3)(2π · 12) = 8π cm ≈ 25.1 cm.

Don't confuse: The arclength depends on both the angle and the radius—same angle on a bigger circle gives a longer arc.

🌟 Defining radians

🌟 The radian measure

The excerpt rewrites the arclength formula by grouping the angle part:

• Old form: Arclength = (fraction of one revolution) · (2πr)
• New form: Arclength = (fraction of one revolution × 2π) · r

Radian measure: The radian measure of an angle is (fraction of one revolution × 2π).

• One complete revolution = 1 × 2π = 2π radians.
• One-quarter revolution = (1/4) × 2π = π/2 radians.
• One-half revolution = (1/2) × 2π = π radians.

🌟 Benchmark angles

The excerpt emphasizes memorizing these:

Degrees	Exact radians	Decimal approx
0°	0	0
90°	π/2	1.57
180°	π	3.14
270°	3π/2	4.71
360°	2π	6.28

Example from excerpt: An angle of 2 radians lies in the second quadrant (between 1.57 and 3.14); 5 radians lies in the fourth quadrant (between 4.71 and 6.28).

🌟 The size of one radian

• 1 radian ≈ 57.3°.
• This is much larger than 1°, which makes sense: there are only about 6.28 radians in a full circle.
• Key property: An angle of 1 radian spans an arc equal to the radius of the circle.

🔄 Converting between degrees and radians

🔄 The conversion factor

Since 360° = 2π radians, dividing both sides by 360° gives:

• 1 = (2π radians) / 360° = (π radians) / 180°
• Also: 1 = 180° / (π radians)

Unit conversion for angles: (180°) / (π radians) = 1

🔄 How to convert

• Radians to degrees: Multiply by 180° / π.
• Degrees to radians: Multiply by π / 180°.

Example from excerpt:

• Convert 3 radians to degrees: 3 × (180°/π) = 540°/π ≈ 171.9°.
• Convert 3° to radians: 3° × (π/180°) = π/60 ≈ 0.05 radians.

Another example:

• Convert 60° to radians: 60° × (π/180°) = π/3 ≈ 1.047 radians.
• Convert 3π/4 radians to degrees: (3π/4) × (180°/π) = 135°.

Don't confuse: The conversion factor flips depending on direction—check which unit you want to eliminate.

🎯 The simplified arclength formula

🎯 When angle is in radians

Recall: Arclength = (fraction of one revolution × 2π) · r

The quantity in parentheses is just the radian measure θ, so:

Arclength formula: s = rθ (where θ is in radians)

• This is much simpler than the degree version.
• Key insight: An angle of 1 radian spans an arc equal to the radius.

🎯 Using the formula

Example from excerpt: Arclength for 2.5 radians on a circle of radius 6 inches:

• s = rθ = 6 × 2.5 = 15 inches.

Rearranged form: θ = s/r (to find the angle from arclength and radius).

Real-world example: Earth's radius is 3960 miles. If you travel 500 miles due north, how many degrees of latitude?

• s = rθ → 500 = 3960θ → θ = 500/3960 ≈ 0.1263 radians.
• Convert to degrees: 0.1263 × (180°/π) ≈ 7.23°.

Another scenario: A clock face has 5 feet of arc from 2 to 3 (which is 1/12 of the circle, or π/6 radians).

• s = rθ → 5 = r(π/6) → r = 30/π ≈ 9.55 feet.

⭕ The unit circle

⭕ Special property

Unit circle: A circle of radius 1.

On a unit circle, r = 1, so the arclength formula becomes s = θ.

Key insight: On a unit circle, the measure of a positive angle in radians equals the length of the arc it spans.

• Walk 1 unit around the circle → you've turned through 1 radian.
• Walk 4 units → you've turned through 4 radians (middle of third quadrant).

⭕ Practical example

From excerpt: You walk around a circular pond of radius 1 mile. After 4 miles, where are you?

• The pond is a unit circle, so you've traversed 4 radians.
• 4 radians is between π (3.14) and 3π/2 (4.71), so you're in the third quadrant, more than halfway but less than three-quarters around.

Another example: An ant walks around a birdbath of diameter 2 feet (radius 1 foot). How far has it walked at 210°?

• Convert 210° to radians: 210° × (π/180°) = 7π/6 radians.
• s = rθ = 1 × (7π/6) = 7π/6 ≈ 3.67 feet.

Don't confuse: The unit circle property (s = θ) only works when r = 1; for other circles, you must use s = rθ.

📊 Summary concepts

📊 Core relationships

1. Arclength depends on two variables: the radius r and the angle θ.
2. Radian definition: (fraction of one revolution × 2π).
3. Conversion factor: 180° / (π radians) = 1.
4. Arclength formula: s = rθ (θ in radians).
5. Unit circle: Arc length equals radian measure (s = θ when r = 1).

📊 Why radians are "natural"

• They eliminate the arbitrary 360 and connect directly to π, which is fundamental to circles.
• The arclength formula is simpler: just multiply radius by angle (no extra fractions).
• Trigonometric and calculus formulas are more concise in radians.

🧭 Overview

🧠 One-sentence thesis

Trigonometric functions can be defined for real numbers (not just angles) by treating radian measure as a dimensionless quantity, which allows us to model periodic phenomena as functions of time or any variable.

📌 Key points (3–5)

• Radians as real numbers: Because radian measure is a ratio of lengths (θ = s/r), it has no units and is just a real number.
• Circular functions definition: To find sin(t) or cos(t) for a real number t, find the sine or cosine of the angle θ = t radians.
• Unit circle coordinates: On a unit circle, the terminal point of an arc of length t has coordinates (cos(t), sin(t)).
• Common confusion: There is no practical difference between "the sine of the number 2" and "the sine of 2 radians"—both use the same calculator operation in radian mode.
• Why it matters: Defining trig functions of real numbers (not just angles) lets us model periodic behavior as functions of time or any other variable.

📐 Trigonometric functions in radian mode

📐 Same angle, different units

• The sine, cosine, or tangent of a particular angle is the same whether the angle is measured in radians or in degrees.
• Example: π/3 radians equals 60°, so sin(π/3) = sin(60°) = √3/2.

🔧 Using a calculator

• Switch your calculator from Degree mode to Radian mode.
• Enter sin(π ÷ 3) to get approximately 0.8660, which is the decimal form of √3/2.
• For angles given as decimals (e.g., 3.5), the calculator will return the trig value directly.
• Example: cos(3.5) ≈ −0.9365 and sin(3.5) ≈ −0.3508 (3.5 radians is a third-quadrant angle).

🔄 Coterminal angles

• Adding or subtracting a multiple of 2π to any angle produces a coterminal angle with the same trig values.
• Example: 5π/2 = 2π + π/2, so sin(5π/2) = sin(π/2) = 1 and cos(5π/2) = cos(π/2) = 0.

🔢 From angles to real numbers

🔢 Radians are dimensionless

Radian measure is a ratio of two lengths (θ = s/r), so the units cancel and radian measure is just a real number.

• This is why we can omit the tag "radians" and assume any angle given without units is in radians.
• Example: The angle 2 is simply the real number 2, which corresponds to an arc of length 2 on a unit circle.

🌐 Sine and cosine of real numbers

We define the trigonometric functions of the real number t by cos(t) = cos(θ) and sin(t) = sin(θ), where θ = t is an angle measured in radians.

• To find the sine or cosine of a real number t, find the sine or cosine of the angle θ = t radians.
• On a unit circle, the measure of a positive angle in radians equals the length of the arc it spans.
• Example: To find cos(2) and sin(2), set your calculator in radian mode and evaluate: cos(2) ≈ −0.4161 and sin(2) ≈ 0.9093.

📊 Modeling periodic phenomena

• Because trig functions are now functions of real numbers, we can model periodic behavior as functions of time or any variable.
• Example: Sunset time in Stockholm on the nth day of the year is T = 3.11 sin(0.017n − 1.38) + 6.03 hours after noon. On January 1 (n = 1), T ≈ 2.99 hours (about 2:59 pm). On July 1 (n = 182), T ≈ 9.11 hours (about 9:07 pm).
• Example: The magnitude of the variable star T Herculis is M = 10.2 − 2.2 cos(0.038t), where t is days after December 27, 2004.

⭕ The circular functions

⭕ Definition on the unit circle

Let P be the terminal point of an arc of length t in standard position on a unit circle. The circular functions of t are defined by:

• cos(t) = x
• sin(t) = y
• tan(t) = y/x, where x ≠ 0

• The point P has coordinates (x, y) on the unit circle.
• This definition agrees with the earlier definition for angles, because P lies on the terminal side of the angle θ = t radians.

🧮 Computing tangent from coordinates

• If P has coordinates (x, y), then tan(t) = y/x.
• Example: For t = 2, the terminal point is approximately (−0.4161, 0.9093), so tan(2) = 0.9093 / (−0.4161) ≈ −2.1853.
• You can verify this by setting your calculator in radian mode and computing tan(2) directly.

📍 Coordinates on a unit circle

The coordinates of the point P determined by an arc of length t in standard position on a unit circle are (x, y) = (cos(t), sin(t)).

• This is the key relationship: the x-coordinate is the cosine, and the y-coordinate is the sine.
• Example: For an arc of length 2.5, the terminal point P is approximately (−0.8, 0.6), so cos(2.5) ≈ −0.8 and sin(2.5) ≈ 0.6.

⭐ Special angles in radians

⭐ The special values table

Degrees	Radians	Sine	Cosine	Tangent
0°	0	0	1	0
30°	π/6	1/2	√3/2	1/√3
45°	π/4	1/√2	1/√2	1
60°	π/3	√3/2	1/2	√3
90°	π/2	1	0	undefined

• These values come from the two special right triangles: the 45-45-90 triangle (sides 1, 1, √2) and the 30-60-90 triangle (sides 1, √3, 2).
• Memorize these benchmarks for mental calculation and estimation.

🔍 Reference angles in radians

Reference angles in radians use the same formulas as in degrees, but replace 180° with π:

• First quadrant: reference angle = θ
• Second quadrant: reference angle = π − θ
• Third quadrant: reference angle = θ − π
• Fourth quadrant: reference angle = 2π − θ

• Example: For θ = 2π/3 (second quadrant), the reference angle is π − 2π/3 = π/3. Since tangent is negative in the second quadrant, tan(2π/3) = −tan(π/3) = −√3.
• Example: For θ = 5π/4 (third quadrant), the reference angle is 5π/4 − π = π/4. Since cosine is negative in the third quadrant, cos(5π/4) = −cos(π/4) = −1/√2.

🧭 Finding terminal points

• To find the coordinates of the terminal point P of an arc of length t, use (cos(t), sin(t)).
• Example: For t = 5π/6, the coordinates are (cos(5π/6), sin(5π/6)) = (−√3/2, 1/2).
• Don't confuse: The expression cos(5π/4) = −1/√2 may look complex, but it just says "the cosine of an angle of about 3.9 radians is about −0.7."

🧭 Overview

🧠 One-sentence thesis

Circular functions (sine, cosine, tangent) can be graphed with the horizontal axis scaled in radians instead of degrees, enabling us to model periodic phenomena as functions of time and solve trigonometric equations graphically or algebraically.

📌 Key points (3–5)

• Scaling difference: The only difference between graphing circular functions and trigonometric functions of angles in degrees is that the horizontal axis is scaled in radians (often in multiples of π).
• Special values as guide points: Using special angle values (multiples of π/12, π/6, π/4, etc.) provides exact coordinates for plotting; quadrantal angles (0, π/2, π, 3π/2, 2π) serve as sufficient guide points for sketching.
• Domain and range: Sine and cosine have domain = all real numbers and range = [-1, 1]; tangent has domain = all real numbers except odd multiples of π/2 and range = all real numbers.
• Common confusion: Radian measure vs. degree measure—the value of sin(t) or cos(t) is the same whether t is measured in radians or degrees, but the scale on the horizontal axis changes; for example, one complete cycle occurs at 2π ≈ 6.28 (not 360).
• Applications: Circular functions model real-world periodic phenomena (Ferris wheels, tides, daylight hours, oscillating springs) as functions of time, and equations can be solved graphically (reading intersection points) or algebraically (using inverse functions and reference angles).

📊 Graphing sine and cosine

📊 Building the sine graph

The graph of y = sin(t) completes one cycle at t = 2π radians (approximately 6.28), reaches maximum y = 1 at t = π/2 (approximately 1.57), and has y = 0 at t = π (approximately 3.14).

How to construct the graph:

• Make a table of values using special angles in radians (0, π/6, π/4, π/3, π/2, etc.).
• Scale the horizontal axis in multiples of π/12 (since all special values are multiples of π/12).
• In quadrants I and II, sine values are positive; in quadrants III and IV, sine values are negative.
• The graph repeats every 2π units (periodic behavior).

Example from the excerpt: At t = 0, sin(0) = 0; at t = π/6, sin(π/6) = 0.500; at t = π/2, sin(π/2) = 1.000; at t = π, sin(π) = 0; at t = 7π/6, sin(7π/6) = -0.500.

📊 Building the cosine graph

The cosine graph is constructed similarly:

• Use the same special angle values.
• At t = 0, cos(0) = 1; at t = π/2, cos(π/2) = 0; at t = π, cos(π) = -1; at t = 3π/2, cos(3π/2) = 0; at t = 2π, cos(2π) = 1.
• The graph also has period 2π, amplitude 1, and midline y = 0.

🎯 Guide points shortcut

Why it works: Because we know the basic shapes of sine and cosine, we can plot only the quadrantal angles (0, π/2, π, 3π/2, 2π) and draw a smooth curve through them.

t	0	π/2	π	3π/2	2π
sin(t)	0	1	0	-1	0
cos(t)	1	0	-1	0	1

📈 Graphing the tangent function

📈 Tangent graph characteristics

The tangent function y = tan(t) has vertical asymptotes at t = π/2 and t = 3π/2 (and at all odd multiples of π/2).

Key differences from sine/cosine:

• The graph has vertical asymptotes where tangent is undefined.
• The period is π (not 2π)—the pattern repeats every π units.
• The range is all real numbers (not bounded between -1 and 1).

Example from the excerpt: Between 0 and π, tan(0) = 0, tan(π/6) ≈ 0.577, tan(π/4) = 1, tan(π/3) ≈ 1.732, tan(π/2) is undefined, tan(3π/4) = -1, tan(5π/6) ≈ -0.577, tan(π) = 0.

📈 Asymptote locations

• In the radian-scaled graph, vertical asymptotes occur at t = π/2, 3π/2, 5π/2, etc. (compare to 90°, 270° in the degree version).
• The graph approaches +∞ from the left of each asymptote and -∞ from the right.

🔍 Solving trigonometric equations

🔍 Graphical solution method

How to solve graphically:

1. Graph the circular function (e.g., y = cos(t)).
2. Graph the horizontal line y = k (the constant on the right side of the equation).
3. Find the t-coordinates of intersection points.

Example from the excerpt: To solve cos(t) = -0.62 between 0 and 2π, graph y = cos(t) and y = -0.62; the intersections appear at approximately t = 2.25 and t = 4.05.

Limitation: Reading from a graph gives approximate solutions; accuracy is limited by the graph's resolution.

🔍 Algebraic solution method

Steps for solving cos(t) = -0.62 algebraically (0 ≤ t ≤ 2π):

1. Recognize that cosine is negative in quadrants II and III (between t = 1.57 and t = 4.71).
2. Use a calculator: cos⁻¹(-0.62) ≈ 2.24 (the second-quadrant angle).
3. Find the reference angle: π - 2.24 = 0.90.
4. Find the third-quadrant angle: π + 0.90 = 4.04.
5. Solutions: t₁ = 2.24 and t₂ = 4.04 radians.

For all solutions (not just 0 to 2π): Add or subtract multiples of 2π to each solution: t = 2.24 ± 2πk and t = 4.04 ± 2πk, where k is any integer.

🔍 Exact solutions with special values

When the equation involves special values, no calculator is needed.

Example from the excerpt: Solve tan(t) = 1.

• Since tan(π/4) = 1, one solution is t₁ = π/4.
• Tangent is also positive in quadrant III: t₂ = π + π/4 = 5π/4.
• All solutions: t = π/4 ± 2πk and t = 5π/4 ± 2πk.

Don't confuse: For tangent, the period is π (not 2π), so you could also write solutions as t = π/4 ± πk.

🌊 Modeling periodic phenomena

🌊 Real-world applications

Circular functions describe periodic phenomena as functions of time (or other variables besides angles).

Example from the excerpt: Ferris wheel

• Radius: 100 feet, rotates once every 8 minutes.
• Height function: h = f(t) = 100 - 100cos(πt/4), where t is minutes after boarding at the bottom.
• After 2 minutes: f(2) = 100 - 100cos(π/2) = 100 - 0 = 100 feet.
• After 4 minutes: f(4) = 100 - 100cos(π) = 100 - 100(-1) = 200 feet.

🌊 Reading the model parameters

From the Ferris wheel graph:

• Midline: h = 100 (average height).
• Amplitude: 100 (distance from midline to maximum or minimum).
• Period: 8 minutes (time for one complete rotation).

Example from the excerpt: Daylight in Glasgow

• Formula: D(t) = 12.25 - 5.25cos(πt/6), where t is months after January 1.
• Midline: D = 12.25 hours.
• Amplitude: 5.25 hours.
• Period: 12 months.
• Longest day: 12.25 + 5.25 = 17.5 hours.
• Less than 8 hours of light: January, February, and late December.

🌊 Interpreting the formula

General form: y = midline ± amplitude · trig function(coefficient · t).

• The coefficient inside the trig function affects the period.
• The ± sign determines whether the function starts at a maximum or minimum.

📐 Domain and range of functions

📐 What domain and range mean

Domain: The set of all possible input values for a function.
Range: The set of all output values for a function.

How to see domain and range in a graph:

• Domain: The set of all x-coordinates of points on the graph (horizontal extent).
• Range: The set of all y-coordinates of points on the graph (vertical extent).

Example from the excerpt: For f(x) = 2x + 5 (a line), the graph extends infinitely in all directions, so domain = all real numbers and range = all real numbers.

📐 Restrictions on domain

Some functions exclude certain input values.

Example from the excerpt: h(x) = √(x + 3)

• Cannot take the square root of a negative number.
• Require x + 3 ≥ 0, so x ≥ -3.
• Domain: x ≥ -3; Range: y ≥ 0.

Example from the excerpt: g(x) = 1/(x - 3)

• Cannot divide by zero, so x ≠ 3.
• Domain: all real numbers except 3.
• The graph has a horizontal asymptote at y = 0, and no input produces output 0.
• Range: all real numbers except 0.

📐 Domain and range of circular functions

Function	Domain	Range	Notes
sin(x)	All real numbers	[-1, 1]	Output bounded by unit circle coordinates
cos(x)	All real numbers	[-1, 1]	Output bounded by unit circle coordinates
tan(x)	All real numbers except odd multiples of π/2	All real numbers	Undefined where cos(x) = 0; unbounded output

Why these ranges: Sine and cosine are defined by coordinates of points on a unit circle, so their values cannot exceed 1 or be less than -1. Tangent = sin/cos increases without bound as it approaches its asymptotes.

🧭 Overview

🧠 One-sentence thesis

The amplitude, period, and midline parameters in sinusoidal functions (sine, cosine, and tangent) control vertical stretching, horizontal compression, and vertical shifting, allowing us to model real-world periodic phenomena like sound waves, tides, and blood pressure.

📌 Key points (3–5)

• Amplitude (|A|) controls vertical stretch/compression: larger |A| means taller waves; it measures the distance from midline to maximum.
• Period (2π/|B|) controls horizontal stretch/compression: larger B compresses the graph horizontally, creating more cycles in the same interval.
• Midline (y = k) shifts the entire graph vertically: the horizontal line around which the function oscillates.
• Common confusion: Amplitude is always nonnegative (use absolute value); a negative coefficient reflects the graph but doesn't change amplitude.
• Choosing sine vs. cosine: both model the same periodic behavior, but the starting point matters—sine starts at the midline, cosine starts at a maximum or minimum.

📐 Amplitude transformations

📏 What amplitude means

Amplitude: The value |A| in y = A sin(x) or y = A cos(x); it measures the vertical distance from the midline to the peak (or trough).

• Amplitude is always nonnegative—take the absolute value of A.
• Larger |A| → taller waves; smaller |A| → flatter waves.
• Example: y = 2 sin(x) has amplitude 2; y = 0.5 sin(x) has amplitude 0.5.

🔄 Negative coefficients

• A negative A reflects the graph across the x-axis but does not change the amplitude.
• Example: both y = 3 cos(x) and y = −3 cos(x) have amplitude 3; the second graph is flipped upside down.
• Don't confuse: "negative amplitude" is not a thing—amplitude is |A|, always positive.

⏱️ Period transformations

🔁 What period means

Period: The horizontal length of one complete cycle; for y = sin(Bx) or y = cos(Bx), the period is 2π / |B|.

• Larger B → shorter period → more cycles in the same interval (horizontal compression).
• Smaller B → longer period → fewer cycles (horizontal stretch).
• Example: y = cos(2x) completes two cycles from 0 to 2π, so period = 2π/2 = π.
• Example: y = cos(x/3) completes one cycle from 0 to 6π, so period = 2π/(1/3) = 6π.

🧮 How to find the period

• Formula: period = 2π / |B|.
• If B = 3, period = 2π/3; if B = 1/4, period = 8π.
• Don't confuse: the coefficient B is inside the function argument, not outside.

📊 Midline transformations

📍 What midline means

Midline: The horizontal line y = k around which the sinusoidal function oscillates; for y = k + sin(x) or y = k + cos(x), the midline is y = k.

• Positive k shifts the graph up; negative k shifts it down.
• Example: y = 2 + sin(x) has midline y = 2; every point is 2 units higher than on y = sin(x).
• Example: y = −3 + cos(x) has midline y = −3; the graph is shifted down 3 units.

🧭 Finding midline from max and min

• If a sinusoidal function has maximum M and minimum m, the midline is y = (M + m)/2.
• The amplitude is then (M − m)/2 = distance from midline to peak.
• Example: blood pressure varies between 70 and 110 → midline = (70 + 110)/2 = 90, amplitude = 110 − 90 = 20.

🎨 Sketching sinusoidal graphs

🗂️ Table-of-values method

• Choose convenient input values for the trig function: multiples of π/4 or π/6 (quadrantal angles).
• Work backwards to find the actual input variable t, and forwards to find the output y.
• Example for y = 2 + 3 cos(4t):
  • Let 4t = 0, π/2, π, 3π/2, 2π → solve for t → t = 0, π/8, π/4, 3π/8, π/2.
  • Compute cos(4t), then 3 cos(4t), then 2 + 3 cos(4t).
• Plot these "guidepoints" and sketch a smooth sinusoidal curve through them.

🎯 Choosing sine vs. cosine

• Sine starts at the midline (y = 0 when x = 0 for y = sin(x)).
• Cosine starts at a maximum (y = 1 when x = 0 for y = cos(x)) or minimum (y = −1 for y = −cos(x)).
• Example: if a graph starts on the midline and increases, use y = A sin(Bx); if it starts on the midline and decreases, use y = −A sin(Bx).
• Example: if a graph starts at a peak, use y = A cos(Bx); if it starts at a trough, use y = −A cos(Bx).

🌊 Modeling real-world phenomena

🩺 Blood pressure example

• Blood pressure varies between 70 and 110 mm Hg, with 60 beats per minute.
• Midline: (70 + 110)/2 = 90; amplitude: 110 − 90 = 20.
• Period: 1/60 minute (one heartbeat) → B = 2π / (1/60) = 120π.
• Model: y = 90 + 20 sin(120πt).

🌊 Tides, sound, and other cycles

• Sinusoidal functions model tides (water level vs. time), sound waves (pressure vs. time), daylight hours (hours vs. month), and more.
• To write a model:
  1. Find max and min → compute midline and amplitude.
  2. Find the period (time for one cycle) → compute B = 2π / period.
  3. Choose sine or cosine based on the starting point.
• Example: daylight varies from 9.6 to 14.4 hours over 12 months → midline = 12, amplitude = 2.4, period = 12 months.

📐 Tangent transformations

📈 Tangent function differences

• The tangent function y = tan(x) does not have an amplitude (it goes to ±∞).
• It does have a period: for y = tan(Bx), period = π / |B| (note: π, not 2π).
• Vertical stretch: the coefficient A in y = A tan(Bx) stretches the graph vertically; compare values at guidepoints (e.g., x = π/4).
• Midline: y = k in y = k + A tan(Bx) shifts the graph up or down.

🔍 Example: y = 1 + 3 tan(2x)

• Midline: y = 1 (shifted up 1 unit).
• Period: π / 2 (compressed horizontally by factor of 2).
• Vertical stretch: factor of 3 (compare to y = tan(x) at guidepoints).
• Don't confuse: tangent period is π/|B|, not 2π/|B|.

📋 Summary table

Parameter	Form	Effect	Formula
Amplitude	y = A sin(x) or y = A cos(x)	Vertical stretch/compression	Amplitude = |A|
Period	y = sin(Bx) or y = cos(Bx)	Horizontal stretch/compression	Period = 2π / |B|
Midline	y = k + sin(x) or y = k + cos(x)	Vertical shift	Midline: y = k
Tangent period	y = tan(Bx)	Horizontal stretch/compression	Period = π / |B|

🧭 Overview

🧠 One-sentence thesis

Horizontal shifts complete the set of transformations needed to model any sinusoidal function, and writing functions in standard form allows us to read all transformation parameters directly from the formula.

📌 Key points (3–5)

• Horizontal shifts: replacing x with (x − h) shifts the graph h units right (if h > 0) or left (if h < 0).
• Order of transformations matters: applying a horizontal compression before a shift produces a different result than shifting first, then compressing.
• Standard form reveals all transformations: y = A sin(B(x − h)) + k or y = A cos(B(x − h)) + k encodes amplitude |A|, period 2π/|B|, midline y = k, and horizontal shift h.
• Common confusion: sin(2x − π/3) is not the same as sin[2(x − π/3)]; factor the input to identify the true horizontal shift.
• Multiple formulas fit the same graph: a sinusoidal curve can be written as a shifted sine or a shifted cosine; choose the form with the smallest shift.

🔄 Horizontal shifts of sine and cosine

🔄 What a horizontal shift does

Horizontal shift: The graphs of y = sin(x − h) and y = cos(x − h) are shifted horizontally compared to y = sin(x) and y = cos(x).

• If h > 0, the graph shifts h units to the right.
• If h < 0, the graph shifts h units to the left.
• The amplitude, midline, and period remain unchanged; only the horizontal position changes.

Example from the excerpt: The graph of g(x) = sin(x − π/4) has the same shape as f(x) = sin(x), but every y-value appears π/4 units farther to the right.

📊 Using a table to see the shift

The excerpt shows a table comparing f(x) = sin(x) and g(x) = sin(x − π/4):

x	0	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
f(x)	0	√2/2	1	√2/2	0	−√2/2	−1	−√2/2	0
g(x)	−√2/2	0	√2/2	1	√2/2	0	−√2/2	−1	−√2/2

• Each function value in g appears π/4 units to the right of where it appears in f.
• This pattern holds for the entire graph.

📝 Sketching shifted graphs with a table

To sketch y = −2 cos(x + π/3):

1. Rewrite as y = −2 cos[x − (−π/3)] to see h = −π/3 (shift π/3 units left).
2. Choose convenient values for the input of cosine: x + π/3 = 0, π/2, π, 3π/2, 2π.
3. Work backwards to find x (subtract π/3 from each input value).
4. Work forwards to find y (evaluate the cosine, then multiply by −2).
5. Plot the guide points and sketch the curve.

Don't confuse: The shift direction depends on the sign of h in (x − h), not on whether you see a plus or minus in the original formula.

⚙️ Order of transformations

⚙️ Why order matters

The excerpt compares two functions:

• f(x) = sin[2(x − π/3)]
• g(x) = sin(2x − π/3)

Both involve a horizontal compression (factor of 2) and a horizontal shift, but:

• f shifts π/3 units to the right.
• g shifts only π/6 units to the right.

🔢 Step-by-step for f(x) = sin[2(x − π/3)]

1. First, replace x with 2x → y = sin(2x) (horizontal compression by factor of 2).
2. Then, replace x with x − π/3 → f(x) = sin[2(x − π/3)] (shift π/3 units right).

The shift happens after the compression, so the full π/3 shift is preserved.

🔢 Step-by-step for g(x) = sin(2x − π/3)

1. First, replace x with x − π/3 → y = sin(x − π/3) (shift π/3 units right).
2. Then, replace x with 2x → g(x) = sin(2x − π/3) (horizontal compression by factor of 2).

The compression happens after the shift, so the shift is also compressed: π/3 ÷ 2 = π/6.

🔧 Factoring to reveal the true shift

To analyze g(x) = sin(2x − π/3) correctly, factor the input:

g(x) = sin(2x − π/3) = sin[2(x − π/6)]

Now you can see B = 2 and h = π/6 directly.

Key principle: Always factor the input into the form B(x − h) to identify the horizontal shift.

📐 Standard form for sinusoidal functions

📐 The standard form

Standard form: y = A sin(B(x − h)) + k or y = A cos(B(x − h)) + k

From this form, read off:

1. Amplitude: |A|
2. Midline: y = k
3. Period: 2π / |B| (where B ≠ 0)
4. Horizontal shift: h units right if h > 0, left if h < 0

🧮 Example: analyzing a function in standard form

Given f(x) = 4 cos(3x + π) − 2:

1. Factor the input: 3x + π = 3(x + π/3), so f(x) = 4 cos[3(x + π/3)] − 2.
2. Identify: A = 4, B = 3, h = −π/3, k = −2.
3. Read the transformations:
   • Amplitude: 4
   • Midline: y = −2
   • Period: 2π / 3
   • Horizontal shift: π/3 units to the left (because h = −π/3 < 0)

📊 Comparison table

Parameter	Formula component	What it controls
Amplitude	|A|	Vertical stretch/compression; distance from midline to max/min
Midline	k	Vertical shift; the horizontal line halfway between max and min
Period	2π / |B|	Horizontal stretch/compression; length of one complete cycle
Horizontal shift	h	Left/right translation; where the cycle starts

Don't confuse: The period depends on B, but the horizontal shift depends on h. Changing B affects both the period and the apparent shift if the input is not factored.

🌍 Modeling with sinusoidal functions

🌍 Finding a formula from a graph

The excerpt shows a graph with:

• Midline: y = 0
• Amplitude: 4
• Period: 2π
• Maximum at x = π/6 (instead of x = 0)

Steps:

1. The graph has a cosine shape (starts at a maximum).
2. The maximum is shifted π/6 units to the right, so h = π/6.
3. Use standard form: y = A cos(B(x − h)) + k.
4. Substitute: A = 4, B = 1 (because period = 2π / B = 2π), h = π/6, k = 0.
5. Formula: y = 4 cos(x − π/6).

Alternative formula: The same graph can be written as y = 4 sin(x + π/3) (a sine function shifted π/3 units left). Both formulas produce identical graphs.

Tip from the excerpt: Choose a formula with a small horizontal shift for simplicity.

🌞 Real-world example: sunspot cycles

The excerpt models sunspot data with a sinusoidal function:

• Period: 10.8 years (given)
• Midline: y = 75 (halfway between max and min)
• Amplitude: 75 (distance from midline to max or min)
• Shape: resembles y = −cos(x) (starts at a minimum)

Initial formula: y = 75 − 75 cos[0.58(x − h)], where B = 2π / 10.8 ≈ 0.58.

Adjusting the fit: By trying different values of h, the excerpt settles on h = 0.3 to better match the data. This shifts the curve slightly to the left.

Key insight: Real-world data rarely fit a perfect sinusoidal curve, but you can adjust the horizontal shift to improve the approximation.

🔍 Steps to model periodic phenomena

1. Identify the midline: average of maximum and minimum values.
2. Find the amplitude: half the distance between max and min.
3. Determine the period: time (or distance) for one complete cycle.
4. Choose sine or cosine: based on where the cycle starts (midline, max, or min).
5. Estimate the horizontal shift: where does the chosen function need to start?
6. Write the formula in standard form.
7. Refine by adjusting h to better fit the data.

Example scenario: Modeling daylight hours in a city. The minimum (9.6 hours) occurs at the winter solstice (t = 0), and the maximum (14.4 hours) occurs at the summer solstice. The period is 12 months. Use a cosine function with a minimum at t = 0, which corresponds to y = −cos(x).

Don't confuse: The period in the formula (2π / B) is in radians, but real-world periods (like months or years) must be converted by setting 2π / B equal to the real-world period.

🧭 Overview

🧠 One-sentence thesis

Trigonometric equations can be systematically solved by finding reference solutions with a calculator and then using symmetry properties of sine, cosine, and tangent to locate all solutions within a given interval.

📌 Key points (3–5)

• Basic solution patterns: Each trig function has a characteristic symmetry—sine and cosine each yield two solutions per cycle, while tangent yields one solution per cycle.
• Multiple solutions from period: Equations like sin(nx) = k have 2n solutions between 0 and 2π because the graph completes n cycles.
• Substitution technique: For equations of the form sin(Bx + C) = k, substitute θ = Bx + C to reduce the problem to a simpler form, then solve for x.
• Common confusion: Don't forget to adjust calculator outputs—if the calculator gives a negative angle or one outside [0, 2π], find a coterminal angle by adding 2π (or π for tangent).
• Why it matters: These techniques apply to real-world periodic models like pistons, tides, Ferris wheels, and moon phases.

🔍 Finding basic solutions

🔍 Sine equations: sin(θ) = k

The excerpt explains that sin(θ) = k always has two solutions between 0 and 2π (for -1 < k < 1).

How to find them:

• Use the calculator to get θ₁ = sin⁻¹(k)
• The second solution uses symmetry: θ₂ = π - θ₁

Why this works:

• The sine graph is symmetric about π/2
• On the unit circle, two points have the same y-coordinate

Sign matters:

• If k > 0: θ₁ = sin⁻¹(k) and θ₂ = π - θ₁
• If k < 0: The calculator returns a negative angle, so add 2π to get θ₁ = sin⁻¹(k) + 2π, then θ₂ = π - sin⁻¹(k)

Example: To solve sin(θ) = -0.58 for 0 ≤ θ < 2π, the calculator gives sin⁻¹(-0.58) = -0.6187. Add 2π to get θ₁ = 5.6645 (fourth quadrant). The second solution is θ₂ = π - (-0.6187) = 3.7603 (second quadrant).

🔍 Cosine equations: cos(θ) = k

Cosine equations also have two solutions between 0 and 2π.

How to find them:

• θ₁ = cos⁻¹(k)
• θ₂ = 2π - θ₁

Why this works:

• The cosine graph is symmetric about the y-axis
• On the unit circle, two points have the same x-coordinate (one above, one below the x-axis)

This pattern holds whether k is positive or negative.

🔍 Tangent equations: tan(θ) = k

Tangent is simpler—only one solution per cycle.

How to find them:

• If k > 0: θ₁ = tan⁻¹(k) and θ₂ = π + θ₁
• If k < 0: θ₁ = tan⁻¹(k) + π and θ₂ = π + θ₁

Why this works:

• The tangent graph repeats every π (not 2π)
• Each cycle has exactly one solution

Example: To solve tan(θ) = -2.4 for 0 ≤ θ < 2π, calculate tan⁻¹(-2.4) = -1.1760. Add π to get θ₁ = 1.9656, then add π again for θ₂ = 5.1072.

🔄 Multiple solutions from frequency

🔄 Equations with sin(nx) or cos(nx)

When the input is nx instead of x, the graph completes n cycles between 0 and 2π.

Key insight:

If n is a positive integer, sin(nθ) = k and cos(nθ) = k each have 2n solutions between 0 and 2π.

Why: Each cycle contributes two solutions, and there are n cycles.

🔄 Finding all solutions step-by-step

The excerpt demonstrates with sin(2x) = 0.5:

1. Solve as if it were sin(θ) = 0.5: get θ = π/6 and θ = 5π/6
2. Replace θ with 2x: so 2x = π/6 and 2x = 5π/6
3. Solve for x: x = π/12 and x = 5π/12
4. Don't stop! The period of sin(2x) is π (not 2π), so add π to each solution: x = π/12 + π = 13π/12 and x = 5π/12 + π = 17π/12

All four solutions: π/12, 5π/12, 13π/12, 17π/12.

🔄 Tangent with frequency

For tan(nx) = k, there is one solution per cycle.

• The period is π/n
• Find the first solution, then add multiples of π/n

Example: tan(2t) = -1 has period π/2. First solution: 2t = 3π/4, so t₁ = 3π/8. Add π/2 repeatedly: t₂ = 7π/8, t₃ = 11π/8, t₄ = 15π/8.

🔧 Using substitution for complex equations

🔧 When to use substitution

For equations like sin(2x + 1.5) = -0.3, the input is not just nx but Bx + C.

Strategy:

Substitute θ = Bx + C to reduce the equation to sin(θ) = k, solve for θ, then replace θ with Bx + C and solve for x.

🔧 Step-by-step substitution process

For sin(Bx + C) = k or cos(Bx + C) = k:

1. Let θ = Bx + C
2. Find two solutions for sin(θ) = k or cos(θ) = k
3. Replace θ with Bx + C in each solution and solve for x
4. Add multiples of 2π/B to find remaining solutions

For tan(Bx + C) = k:

1. Let θ = Bx + C
2. Find one solution for tan(θ) = k
3. Replace θ with Bx + C and solve for x
4. Add multiples of π/B to find remaining solutions

🔧 Worked example

Solve sin(2x + 1.5) = -0.3 for 0 ≤ x ≤ 2π.

1. Let θ = 2x + 1.5
2. Solve sin(θ) = -0.3: θ₁ = sin⁻¹(-0.3) = -0.3047, so add 2π to get θ₁ = 5.9785; also θ₂ = π - sin⁻¹(-0.3) = 3.4463
3. Replace θ: 2x + 1.5 = 3.4463 gives x₁ = 0.9731; 2x + 1.5 = 5.9785 gives x₂ = 2.2392
4. Period is 2π/2 = π, so add π: x₃ = 4.1147, x₄ = 5.3808

Don't confuse: The period for finding additional solutions is 2π/B (or π/B for tangent), not the original 2π.

🌊 Applications to periodic models

🌊 Piston example

A piston pumps at 1000 cycles per second, moving 16 cm between lowest and highest positions.

Model: h(t) = -8 sin(2000πt) + 8

• Amplitude A = 16/2 = 8
• Midline k = 8
• Period = 1/1000, so B = 2000π
• Negative sine because it starts at midline moving down

Question: When is the piston at 14 cm?

Solve: -8 sin(2000πt) + 8 = 14
→ sin(2000πt) = -3/4
→ Substitute θ = 2000πt, solve sin(θ) = -3/4
→ Get t ≈ 0.000635 and t ≈ 0.000865 seconds

🌊 Key modeling steps

1. Identify amplitude, period, midline from the physical situation
2. Write the sinusoidal function
3. Set the function equal to the target value
4. Use substitution to solve
5. Interpret solutions in context (e.g., time in seconds, position in meters)

📋 Summary table

Equation type	Solutions per cycle	How to find second solution	Period for multiples
sin(θ) = k	2	θ₂ = π - θ₁	2π
cos(θ) = k	2	θ₂ = 2π - θ₁	2π
tan(θ) = k	1	θ₂ = θ₁ + π	π
sin(nx) = k	2n total	Same pattern, then add 2π/n	2π/n
cos(nx) = k	2n total	Same pattern, then add 2π/n	2π/n
tan(nx) = k	n total	Add π/n repeatedly	π/n

🧭 Overview

🧠 One-sentence thesis

The Pythagorean and tangent identities allow us to find any of the three basic trigonometric values (sine, cosine, tangent) when we know just one of them.

📌 Key points (3–5)

• What identities are reviewed: the Pythagorean identity and the tangent identity, which relate sine, cosine, and tangent.
• Core relationship: sine squared plus cosine squared equals 1; tangent equals sine divided by cosine.
• Why they matter: knowing one trigonometric value lets you calculate the other two using these identities.
• Common confusion: these are relationships between the three functions, not definitions of the functions themselves.

🔗 The two fundamental identities

🔗 Pythagorean identity

Pythagorean identity: sin squared of theta plus cos squared of theta equals 1.

• This identity connects sine and cosine for any angle theta.
• It is called "Pythagorean" because it relates to the Pythagorean theorem applied to the unit circle.
• How to use it: if you know sine of an angle, you can solve for cosine (or vice versa).
• Example: if sin(theta) is known, then cos squared(theta) = 1 minus sin squared(theta), so you can find cos(theta) by taking the square root.

🔗 Tangent identity

Tangent identity: tan(theta) equals sin(theta) divided by cos(theta).

• This identity expresses tangent as a ratio of sine to cosine.
• How to use it: if you know both sine and cosine, divide them to get tangent; if you know tangent and one of sine or cosine, you can find the other.
• Example: if sin(theta) and cos(theta) are both known, compute their quotient to find tan(theta).

🧮 Using the identities together

🧮 Finding all three values from one

• The excerpt states: "If we know one of the three trig values for an angle, we can find the other two by using these identities."
• Strategy:
  • Start with the known value (sine, cosine, or tangent).
  • Use the Pythagorean identity to relate sine and cosine.
  • Use the tangent identity to connect tangent with sine and cosine.
• Example: given cos(theta), use the Pythagorean identity to solve for sin(theta), then use the tangent identity to compute tan(theta) as sin(theta) divided by cos(theta).

🧮 Don't confuse relationships with definitions

• These identities are relationships between the three functions, not the original definitions of sine, cosine, or tangent.
• The identities assume you already understand what sine, cosine, and tangent mean (from earlier chapters).
• They provide a way to move between the three values algebraically once you have one of them.

🧭 Overview

🧠 One-sentence thesis

Inverse trigonometric functions allow us to find angles when we know their sine, cosine, or tangent values, but we must restrict the domain of the original trig functions to ensure the inverses are also functions.

📌 Key points (3–5)

• What inverse trig functions do: They reverse the operation of sine, cosine, and tangent—given a ratio, they return an angle.
• Why domain restriction matters: The original trig functions are not one-to-one (they repeat values), so we must restrict their domains to make their inverses qualify as functions.
• Range conventions: Each inverse trig function returns angles only in a specific interval (e.g., arcsin returns angles between negative pi/2 and pi/2).
• Common confusion: Composing a trig function with its inverse in one order always returns the original input (e.g., sin(arcsin(x)) = x), but reversing the order may not (e.g., arcsin(sin(x)) may not equal x).
• Modeling applications: Inverse trig functions model real-world situations where an angle depends on a known ratio (distance, height, etc.).

🔄 Core concept: inverse functions

🔄 What an inverse function does

Two functions are called inverse functions if each "undoes" the results of the other function.

• Example: Cubing and taking cube roots are inverse operations. If f(x) = x³, then f⁻¹(x) = ³√x.
• The table of values for an inverse function swaps the x and y columns of the original function's table.
• To find a formula for the inverse: interchange x and y in the original formula, then solve for y.

⚠️ Notation warning

• The notation f⁻¹(x) does not mean 1/f(x).
• The superscript −1 indicates the inverse function, not a reciprocal.
• Example: If f(x) = x³, then f⁻¹(x) = ³√x, which is not the same as 1/x³.

🪞 Symmetry property

• The graphs of f(x) and f⁻¹(x) are symmetric about the line y = x.
• This symmetry occurs because we interchange x and y to define the inverse.
• The domain of f⁻¹ equals the range of f; the range of f⁻¹ equals the domain of f.

🧪 One-to-one functions and the horizontal line test

🧪 When an inverse is a function

Horizontal Line Test: A function passes the Horizontal Line Test if every horizontal line intersects the graph at most once. In that case, there is only one x-value for each y-value, and the function is called one-to-one.

• A function f has an inverse function if and only if f is one-to-one.
• Not every function has an inverse that is also a function.
• Example: F(x) = x² − 4 is not one-to-one (fails the horizontal line test), so its inverse is not a function.

✂️ Restricting the domain

• Sometimes we restrict the domain of a function to make it one-to-one.
• Example: If we restrict F(x) = x² − 4 to x ≥ 0, the resulting function is one-to-one, and its inverse f⁻¹(x) = √(x + 4) is a function.
• This technique is essential for defining inverse trigonometric functions.

📐 The three inverse trig functions

📐 Inverse sine (arcsin)

sin⁻¹(x) = θ if and only if sin(θ) = x and −π/2 ≤ θ ≤ π/2.

• Domain: −1 ≤ x ≤ 1
• Range: −π/2 ≤ y ≤ π/2 (first and fourth quadrants)
• We restrict the sine function to the interval [−π/2, π/2] to make it one-to-one.
• Positive inputs return first-quadrant angles; negative inputs return fourth-quadrant angles.
• Alternate notation: arcsin(x) or asin(x).

📐 Inverse cosine (arccos)

cos⁻¹(x) = θ if and only if cos(θ) = x and 0 ≤ θ ≤ π.

• Domain: −1 ≤ x ≤ 1
• Range: 0 ≤ y ≤ π (first and second quadrants)
• We restrict the cosine function to the interval [0, π] to make it one-to-one.
• Positive inputs return first-quadrant angles; negative inputs return second-quadrant angles.
• Alternate notation: arccos(x) or acos(x).

📐 Inverse tangent (arctan)

tan⁻¹(x) = θ if and only if tan(θ) = x and −π/2 < θ < π/2.

• Domain: all real numbers
• Range: −π/2 < y < π/2 (first and fourth quadrants)
• We restrict the tangent function to the interval (−π/2, π/2) to make it one-to-one.
• Positive inputs return first-quadrant angles; negative inputs return fourth-quadrant angles.
• Alternate notation: arctan(x) or atan(x).

⚖️ Composition properties

⚖️ When composition returns the original value

The following identities always hold:

• sin(sin⁻¹(x)) = x for −1 ≤ x ≤ 1
• cos(cos⁻¹(x)) = x for −1 ≤ x ≤ 1
• tan(tan⁻¹(x)) = x for all x

Why: Applying an inverse trig function gives an angle in the restricted range, and applying the trig function to that angle returns the original ratio.

⚖️ When composition may not return the original value

The following may not be true:

• sin⁻¹(sin(x)) may not equal x
• cos⁻¹(cos(x)) may not equal x
• tan⁻¹(tan(x)) may not equal x

Why: The inverse trig function returns only one angle (in its restricted range) even though many angles have the same trig value. If the original angle x is outside the restricted range, the inverse function returns a different angle with the same trig value.

Example: sin⁻¹(sin(3π/4)) = sin⁻¹(√2/2) = π/4, not 3π/4, because π/4 is the angle in [−π/2, π/2] whose sine is √2/2.

🛠️ Simplifying expressions

🛠️ Strategy: treat the inverse as an angle

• When simplifying expressions like sin(cos⁻¹(3/5)), let θ = cos⁻¹(3/5).
• This means cos(θ) = 3/5 and θ is in the range of the inverse cosine function.
• Use the Pythagorean identity or a right triangle to find other trig ratios of θ.
• Example: If cos(θ) = 3/5 and θ is in the first quadrant, then sin(θ) = 4/5 (from sin²(θ) + cos²(θ) = 1).

🛠️ Working with variables

• The same technique applies to expressions with variables.
• Example: To simplify tan(sin⁻¹(x)), let θ = sin⁻¹(x), so sin(θ) = x.
• Then cos(θ) = √(1 − x²) (from the Pythagorean identity), and tan(θ) = x/√(1 − x²).

🛠️ Don't confuse with reciprocals

• Remember: sin⁻¹(x) is not the same as 1/sin(x).
• The inverse function notation f⁻¹ is completely different from the reciprocal 1/f.

🌍 Modeling with inverse trig functions

🌍 When to use inverse trig functions

Inverse trig functions model situations where:

• An angle changes as a geometric quantity (distance, height) changes.
• We know a trigonometric ratio and need to express the angle in terms of other variables.

🌍 Example scenario

A 3-meter tall tapestry is at eye level. The angle θ subtended by the tapestry depends on your distance d from the wall.

• From the right triangle: tan(θ) = 3/d, so d = 3/tan(θ).
• To express θ as a function of d: θ = tan⁻¹(3/d).
• This formula gives the viewing angle for any distance d.

🌍 Interpreting results

• Evaluating an inverse trig function at a specific value gives a concrete angle.
• Example: If d = 5, then θ = tan⁻¹(3/5) gives the angle in radians when standing 5 meters from the wall.

🧭 Overview

🧠 One-sentence thesis

The secant, cosecant, and cotangent functions are defined as the reciprocals of cosine, sine, and tangent respectively, and they extend our toolkit for solving trigonometric problems while obeying their own identities and graphical behaviors.

📌 Key points (3–5)

• Three new functions: secant (sec), cosecant (csc), and cotangent (cot) are defined as reciprocals of the basic trig functions.
• Not the same as inverse functions: reciprocal functions like sec(x) are fundamentally different from inverse functions like cos⁻¹(x); one gives a ratio, the other gives an angle.
• Undefined values: each reciprocal function is undefined wherever its denominator (the corresponding basic function) equals zero.
• Common confusion: distinguishing reciprocal from inverse—sec(0.8) equals 1 divided by cos(0.8), while cos⁻¹(0.8) is the angle whose cosine is 0.8.
• Graphical behavior: the graphs have vertical asymptotes where the basic functions cross zero, and their ranges exclude values between -1 and 1 (for sec and csc).

📐 Definitions and basic properties

📐 The three reciprocal functions

Secant: sec(θ) = r/x (where r is the hypotenuse, x is the adjacent side in standard position)
Cosecant: csc(θ) = r/y (where y is the opposite side)
Cotangent: cot(θ) = x/y

In terms of the basic functions:

Reciprocal function	Definition	Relationship
Secant	sec(θ) = 1/cos(θ)	Reciprocal of cosine
Cosecant	csc(θ) = 1/sin(θ)	Reciprocal of sine
Cotangent	cot(θ) = 1/tan(θ)	Reciprocal of tangent

📐 Right triangle ratios

For an acute angle θ in a right triangle:

• sec(θ) = hypotenuse / adjacent
• csc(θ) = hypotenuse / opposite
• cot(θ) = adjacent / opposite

These mirror the basic ratios but with numerator and denominator swapped (compared to cos, sin, and tan respectively).

⚠️ Critical distinction: reciprocal vs. inverse

Don't confuse these two concepts:

• Reciprocal function sec(0.8) = 1/cos(0.8) ≈ 1.4353 (a number, the reciprocal of a ratio)
• Inverse function cos⁻¹(0.8) ≈ 0.6435 radians (an angle whose cosine is 0.8)

The excerpt emphasizes: "The reciprocal functions are not the same as the inverse trig functions!"

Example: sec(x) gives you 1 divided by the cosine of x radians, while cos⁻¹(x) gives you the angle (in a restricted range) whose cosine equals x.

🔢 Evaluating reciprocal functions

🔢 Using a calculator

Calculators lack dedicated keys for sec, csc, and cot. Instead:

• To find sec(47°): enter 1 ÷ cos(47°), or compute cos(47°) first and use the reciprocal key x⁻¹.
• The same method applies to csc and cot.

Example from the excerpt: sec(47°) ≈ 1.466.

🔢 Exact values for special angles

The excerpt provides a table of exact values. Key examples:

θ	sec(θ)	csc(θ)	cot(θ)
0	1	undefined	undefined
π/6	2√3/3	2	√3
π/4	√2	√2	1
π/3	2	2√3/3	1/√3
π/2	undefined	1	0

Note the pattern: sec and csc swap the values of their reciprocal partners at complementary angles.

🔢 Finding all six trig ratios from one

If you know one trig ratio, you can find all others using identities.

Example from the excerpt: If sec(θ) = 3 and -π/2 ≤ θ ≤ 0 (fourth quadrant):

1. sec(θ) = 3 means cos(θ) = 1/3
2. Draw a reference triangle: horizontal leg x = 1, hypotenuse r = 3
3. Use Pythagorean theorem: y = -√8 = -2√2 (negative in fourth quadrant)
4. Now compute all six ratios from x, y, r

This method works because all six ratios are determined by the coordinates (x, y) and distance r.

🚫 Undefined values and domains

🚫 Where each function is undefined

Each reciprocal function is undefined when its denominator equals zero:

• Secant is undefined when cos(θ) = 0 → at odd multiples of 90° (or π/2 radians)
• Cosecant is undefined when sin(θ) = 0 → at multiples of 180° (or π radians)
• Cotangent is undefined when tan(θ) = 0 → at multiples of 180° (or π radians)

Important note from the excerpt: although tan(π/2) is undefined, cot(π/2) = 0. This is because cot relates to the ratio x/y, which equals zero when x = 0 (on the positive y-axis).

🚫 Domain and range

Secant:

• Domain: all real numbers except odd multiples of π/2
• Range: y ≥ 1 or y ≤ -1 (excludes values between -1 and 1)

Cosecant:

• Domain: all real numbers except integer multiples of π
• Range: (-∞, -1] ∪ [1, ∞)

Cotangent:

• Domain: all real numbers except integer multiples of π
• Range: all real numbers

The restricted ranges of sec and csc reflect that cosine and sine are bounded between -1 and 1, so their reciprocals cannot fall between -1 and 1.

📊 Graphs of reciprocal functions

📊 Constructing graphs from basic functions

The excerpt demonstrates constructing y = sec(x) from y = cos(x):

1. Insert vertical asymptotes where cos(x) = 0 (at x = -π/2, π/2, 3π/2, etc.)
2. Plot key points by taking reciprocals: where cos(x) = 1, sec(x) = 1; where cos(x) = -1, sec(x) = -1
3. Fill in the curves: as cos(x) decreases toward 0, sec(x) increases toward ∞

The resulting graph has U-shaped branches between asymptotes, opening upward where cosine is positive and downward where cosine is negative.

📊 Key graphical features

Secant and cosecant:

• Vertical asymptotes where the base function is zero
• No values between -1 and 1
• Period of 2π (same as their base functions)

Cotangent:

• Vertical asymptotes at multiples of π
• Decreasing on each interval between asymptotes
• Range includes all real numbers
• Period of π

Don't confuse: the graphs of y = csc(x) and y = cot(x) have asymptotes at the same x-values (multiples of π), but their behavior between asymptotes differs—csc has U-shaped branches, cot is always decreasing.

🧮 Solving equations and using identities

🧮 Solving equations with reciprocal functions

To solve equations like csc(θ) = k:

1. Take the reciprocal of both sides: sin(θ) = 1/k
2. Solve the resulting equation in the basic function
3. Check the range restrictions

Example from the excerpt: csc(θ) = 2√3/3

• Taking reciprocals: sin(θ) = 3/(2√3) = √3/2
• Solutions: θ = π/3 and θ = 2π/3 (between 0 and 2π)

Note: equations sec(θ) = k or csc(θ) = k have no solution if -1 < k < 1, because the ranges of these functions exclude that interval.

🧮 The cotangent identity

cot(θ) = 1/tan(θ) = cos(θ)/sin(θ), provided sin(θ) ≠ 0

This identity is useful for converting expressions. The excerpt shows it can be derived from the definitions of sine and cosine.

🧮 Two more Pythagorean identities

1 + tan²(θ) = sec²(θ)
1 + cot²(θ) = csc²(θ)

These are derived from the original Pythagorean identity sin²(θ) + cos²(θ) = 1 by dividing through by cos²(θ) or sin²(θ) respectively.

When to use them: These identities are especially useful when you know tan(θ) or cot(θ) and need to find sec(θ) or csc(θ).

Example from the excerpt: If tan(α) = 3/5 and α is in the third quadrant:

1. sec²(α) = 1 + tan²(α) = 1 + 9/25 = 34/25
2. sec(α) = ±√(34/25) = ±√34/5
3. Choose the negative root (both sine and cosine are negative in quadrant III)
4. cos(α) = -5/√34 (reciprocal of secant)

🧮 Simplifying expressions

Strategy: Convert all trig ratios to sines and cosines first, then simplify.

Example from the excerpt: sec(θ) - tan(θ)sin(θ)

1. Replace: 1/cos(θ) - [sin(θ)/cos(θ)]·sin(θ)
2. Combine: [1 - sin²(θ)]/cos(θ)
3. Use Pythagorean identity: cos²(θ)/cos(θ) = cos(θ)

This method works because sine and cosine are the fundamental building blocks; expressing everything in terms of them often reveals simplifications.

🔧 Applications

🔧 Right triangle problems

The excerpt gives a flagpole shadow problem:

• Setup: flagpole height h, sun's angle θ from ground, shadow length L
• Relationship: L/h = cot(θ), so L = h·cot(θ)
• Example: for h = 3 meters and θ = 20°, L = 3·cot(20°) ≈ 8.24 meters

Why cotangent? Because we have adjacent (shadow) over opposite (height), which is the definition of cot(θ).

🔧 When reciprocal functions are more convenient

Although any relationship can be expressed using sin, cos, and tan, sometimes reciprocal functions simplify the formula.

Example from the excerpt: Area of a regular n-sided polygon with perimeter L: A = (L²/4n)·cot(π/n)

Using cotangent here is more compact than writing cos/sin explicitly.

🧭 Overview

🧠 One-sentence thesis

Vectors—quantities defined by both magnitude and direction—can be added geometrically using the parallelogram rule, multiplied by scalars to change their length, and broken into horizontal and vertical components to simplify calculations.

📌 Key points (3–5)

• What a vector is: a mathematical tool indicating both direction and magnitude (size), often represented as an arrow.
• How to add vectors: use the parallelogram rule—place the base of the second vector at the head of the first, then draw the resultant from the base of the first to the head of the second.
• Scalar multiplication: multiplying a vector by a positive scalar changes its length but not direction; a negative scalar reverses direction.
• Components simplify calculations: any vector can be broken into horizontal (x) and vertical (y) components using trigonometry; adding vectors by components avoids the laws of sines and cosines.
• Common confusion: the magnitude of u + v is not simply ||u|| + ||v|| unless the vectors are parallel; vector addition is geometric and depends on the angle between them.

🎯 What vectors represent

🎯 Definition and visual representation

Vector: a mathematical tool that indicates both a direction and a size, or magnitude.

• Vectors are often drawn as arrows.
• The length of the arrow represents the magnitude.
• The direction of the arrow shows the vector's direction.
• Example: wind speed and direction on a weather map; the arrow points where the wind blows, and its length (or color) shows speed.

📍 Position vectors and displacement

Position vector: a vector used to designate location relative to a fixed landmark.

• Example: "The airport is 8 miles northeast of the town hall."
• Both distance (8 miles) and direction (northeast) are needed.

Displacement vector: represents the change in position from one point to another.

• Example: traveling 6 miles east then 8 miles north results in a net displacement of 10 miles at 53.1° north of east.

🌀 Real-world application: hurricanes

• The excerpt introduces vectors through hurricane wind patterns.
• Near the surface, winds spiral toward the eye (center) of a hurricane.
• Four main forces (pressure gradient, Coriolis, centrifugal, friction) act as vectors.
• Wind shear (variation in wind speed or direction over short distances) affects storm development; low shear allows the storm to remain vertically aligned.

📏 Notation and equality

📏 Notation for vectors

• In print: boldface v or u.
• Handwritten: arrow above the variable, ~v.
• Magnitude (length) of vector v is denoted ||v||, which is a scalar (a number, not a vector).

Caution: v denotes a vector; ||v|| denotes a scalar.

⚖️ When two vectors are equal

• Two vectors are equal if and only if they have the same length and the same direction.
• They can start at different locations.
• Example: vectors c and g in the excerpt are equal because they have identical length and direction, even though they are drawn in different places.

Don't confuse: vectors with the same length but different directions are not equal; vectors with the same direction but different lengths are not equal.

✖️ Scalar multiplication

✖️ Multiplying a vector by a number

Scalar multiplication: multiplying a vector v by a real number k produces a new vector kv.

• If k > 0: kv points in the same direction as v, and ||kv|| = k · ||v||.
• If k < 0: kv points in the opposite direction to v, and ||kv|| = |k| · ||v||.
• Example:
  • (2/3)v is two-thirds as long as v and points the same way.
  • −√3 v is about 1.7 times as long as v and points the opposite way.

🔄 Why "scalar"?

• Real numbers are called scalars because they "scale" (resize) vectors.
• Scalar multiplication changes the magnitude (and possibly direction) but preserves the line of action (slope).

➕ Adding vectors: the parallelogram rule

➕ Geometric addition

Resultant vector: the sum of two vectors, obtained by placing the base of the second vector at the head of the first, then drawing a vector from the base of the first to the head of the second.

• Notation: w = u + v.
• The resultant forms the third side of a triangle (or the diagonal of a parallelogram).
• Example: traveling 6 miles east then 8 miles north gives a net displacement of 10 miles at 53.1° north of east.

🔁 Commutativity

• Vector addition is commutative: u + v = v + u.
• You can add in either order; the resultant is the same.
• Visualized as the diagonal of a parallelogram formed by u and v.

⚠️ Common mistake

Caution: Unless u and v are parallel, ||u + v|| ≠ ||u|| + ||v||.

• Vector addition is geometric, not arithmetic.
• The length of the sum depends on the lengths of u and v and the angle between them.
• Example: if u and v point in opposite directions, the resultant can be shorter than either vector.

🧮 Calculating the resultant

• Use the law of cosines to find the magnitude of the resultant.
• Use the law of sines to find the direction.
• Example (from the excerpt): hiking 4 miles southwest then 3 miles 30° east of north results in a displacement of 1.348 miles at 9.8° south of west.

🌊 Velocity as a vector

🌊 Velocity vs. speed

Velocity: the combination of speed and direction of motion (not just speed alone).

• Velocities add like vectors.
• If an object has two simultaneous motions, the resulting displacement is the same as if the motions occurred one after the other.

🪲 Example: beetle on a conveyor belt

• Conveyor belt moves at 4 in/sec; beetle walks perpendicular at 2 in/sec.
• After 1 second, the beetle's actual displacement is √(2² + 4²) ≈ 4.47 inches at 26.6° from the belt's direction.
• The two motions (belt + beetle) can be treated as vectors and added.

🚢 Ship and current

• A ship travels at 15 mph on bearing 280°; current flows at 6 mph on bearing 160°.
• The ship's actual velocity is the vector sum of its velocity relative to the water and the water's velocity.
• Use the law of cosines and sines to find the resultant speed (≈13.1 mph) and bearing (≈256.6°).

✈️ Compensating for wind or current

• Sometimes you know the desired velocity w and one component (e.g., wind u), and you need to find the heading v such that v + u = w.
• Example: Barbara wants to travel west at 15 mph, but there's a current at 3 mph 45° east of north. She should head at 17.25 mph, 8.1° south of west.

Don't confuse: the heading you aim for vs. the actual path you travel; they differ when there's a current or wind.

🧩 Components of a vector

🧩 Breaking a vector into x and y parts

Vector components: the horizontal and vertical vectors v_x and v_y such that v = v_x + v_y.

• Any vector can be decomposed into components along coordinate axes.
• The components (scalars) are:
  • v_x = ||v|| cos(θ)
  • v_y = ||v|| sin(θ) where θ is the angle measured counterclockwise from the positive x-axis.

🔢 Components are scalars

• v_x and v_y are numbers, not vectors.
• They can be positive, negative, or zero, depending on the quadrant.

Quadrant	v_x	v_y
I (0°–90°)	+	+
II (90°–180°)	−	+
III (180°–270°)	−	−
IV (270°–360°)	+	−

🧮 Reconstructing magnitude and direction

• Given components v_x and v_y:
  • Magnitude: ||v|| = √(v_x² + v_y²)
  • Direction: tan(θ) = v_y / v_x
• Caution: there are always two angles with a given tangent; use the signs of v_x and v_y to choose the correct quadrant.

Example: if v_x < 0 and v_y > 0, θ is in the second quadrant (90° < θ < 180°).

🔧 Using components to add vectors

🔧 Component-wise addition

• To add u and v:

  1. Find the components of each: u_x, u_y and v_x, v_y.
  2. Add corresponding components: w_x = u_x + v_x, w_y = u_y + v_y.
  3. Compute the magnitude and direction of w from w_x and w_y.

• Advantage: no need for the law of sines or cosines; only right-triangle trigonometry.

✈️ Example: plane with wind

• Plane flies at 300 mph, 30° north of west (θ = 150°).
  • v_x = 300 cos(150°) ≈ −259.81
  • v_y = 300 sin(150°) = 150
• Wind blows at 40 mph from 10° south of west (i.e., toward 10° north of east).
  • u_x = 40 cos(10°) ≈ 39.30
  • u_y = 40 sin(10°) ≈ 6.95
• Resultant:
  • w_x = 39.30 − 259.81 ≈ −220.51
  • w_y = 6.95 + 150 = 156.95
  • ||w|| = √(220.51² + 156.95²) ≈ 270.66 mph
  • tan(θ) = 156.95 / (−220.51) ≈ −0.7118 → θ ≈ 145° (second quadrant)
• The plane's ground speed is ≈270.66 mph at 35° north of west.

⚠️ Angle ambiguity

Caution: tan⁻¹(−0.7118) ≈ −36°, but that's in the fourth quadrant. Since w_x < 0 and w_y > 0, the correct angle is 180° − 36° = 144° (or use the sketch to confirm the second quadrant).

📚 Summary of operations

📚 Two main operations on vectors

Operation	Description	Effect
Scalar multiplication	Multiply v by scalar k	If k > 0: same direction, magnitude k·
Vector addition	Add u and v using parallelogram rule	Place base of v at head of u; resultant goes from base of u to head of v

📚 Key formulas

• Components from magnitude and direction:

  • v_x = ||v|| cos(θ)
  • v_y = ||v|| sin(θ)

• Magnitude and direction from components:

  • ||v|| = √(v_x² + v_y²)
  • tan(θ) = v_y / v_x (check quadrant!)

• Adding vectors by components:

  • (u + v)_x = u_x + v_x
  • (u + v)_y = u_y + v_y

🔄 Subtraction (briefly mentioned)

• Multiplying v by −1 gives −v, which has the same magnitude but opposite direction.
• Subtraction: u − v = u + (−v).
• To subtract, add the opposite vector.

🧭 Overview

🧠 One-sentence thesis

The coordinate form of a vector—expressed as ai + bj using unit vectors—makes vector addition, scalar multiplication, and problem-solving much more efficient than working with magnitude and direction alone.

📌 Key points (3–5)

• What coordinate form is: expressing any vector as a sum of horizontal and vertical components using unit vectors i (x-direction) and j (y-direction).
• Converting between forms: you can switch from geometric form (magnitude and direction) to coordinate form (components) and back using trigonometry.
• Why coordinate form is powerful: addition and scalar multiplication become simple arithmetic on components, avoiding complex geometric constructions.
• Common confusion: the magnitude of a sum ‖u + v‖ is usually not equal to ‖u‖ + ‖v‖; vector addition is not the same as adding lengths.
• Real applications: coordinate form simplifies navigation, force problems, and any situation involving multiple vector quantities.

📐 What coordinate form means

📐 Unit vectors i and j

A unit vector is a vector of magnitude 1.

• i denotes the unit vector pointing in the positive x-direction.
• j denotes the unit vector pointing in the positive y-direction.
• Any horizontal or vertical vector can be written as a scalar multiple of i or j.
• Example: 4i means "4 units in the x-direction"; 3j means "3 units in the y-direction."

🧩 Definition of coordinate form

Coordinate form of a vector: The vector v = ai + bj is the vector whose horizontal component is a and whose vertical component is b.

• The coefficients a and b align with the coordinate axes.
• Example: v = 4i + 3j means the vector extends 4 units horizontally and 3 units vertically from its base.
• This is also called expressing the vector "in terms of i and j."

🔍 Reading coordinate form from a diagram

• Count how many units the vector moves in the x-direction (positive = right, negative = left) → that's the coefficient of i.
• Count how many units it moves in the y-direction (positive = up, negative = down) → that's the coefficient of j.
• Example: A vector from its base extending 4 units left and 6 units up is v = −4i + 6j.

🔄 Converting between geometric and coordinate forms

🔄 From geometric to coordinate

If a vector has magnitude ‖v‖ and direction angle θ (measured counterclockwise from the positive x-axis), its coordinate form is:

• a = ‖v‖ cos(θ)
• b = ‖v‖ sin(θ)
• So v = ai + bj

Why this works: The components are the legs of a right triangle whose hypotenuse is the vector.

Example: A vector with magnitude 4 and direction 29° becomes w = 3.50i + 1.94j.

🔄 From coordinate to geometric

If v = ai + bj, then:

• Magnitude: ‖v

🧭 Overview

🧠 One-sentence thesis

Vectors can be decomposed into components pointing in arbitrary directions (not just horizontal and vertical), which is useful for analyzing motion and forces in different orientations.

📌 Key points (3–5)

• Beyond horizontal and vertical: vectors can be resolved into components along any chosen directions, not only the standard coordinate axes.
• Real-world motivation: the excerpt introduces this through a physical scenario comparing vertical drop versus motion on an inclined ramp.
• What "components" means: breaking a single vector into parts that point in specified directions.

🧩 Resolving vectors in arbitrary directions

🧩 What component resolution means

Components: the parts of a vector that point in chosen directions.

• Previously, vectors were broken into horizontal (i) and vertical (j) parts.
• The excerpt signals that the same idea applies to any pair of directions, not just perpendicular coordinate axes.
• This flexibility is essential when analyzing motion or forces along surfaces that are tilted or oriented at angles.

🎯 Why non-standard components matter

• The excerpt motivates this with a thought experiment:
  • Delbert's ball: dropped vertically, falls straight down under gravity.
  • Francine's ball: released on an inclined ramp, moves along the ramp's slope.
• Both balls experience the same gravitational force, but the effective component of that force differs depending on the surface orientation.
• Example: On a ramp, gravity's component along the ramp causes the ball to roll; the component perpendicular to the ramp is balanced by the ramp's support.

🔧 Practical implications

🔧 Analyzing motion on inclined surfaces

• When a force (like gravity) acts on an object on a slope, resolving the force into components parallel and perpendicular to the slope clarifies:
  • Which part causes motion along the slope.
  • Which part is counteracted by the surface.
• Don't confuse: the total force magnitude is the same, but its effect depends on the chosen component directions.

---

Note: The excerpt provided is primarily exercise problems from earlier in the chapter, with only a brief introductory paragraph for section 9.3. The substantive content about the dot product itself is not included in this excerpt.

🧭 Overview

🧠 One-sentence thesis

Polar coordinates provide an alternative to Cartesian coordinates by specifying locations through distance from a central pole and angle from a reference axis, which simplifies equations for objects with radial symmetry.

📌 Key points (3–5)

• What polar coordinates measure: distance from the pole (r) and angle from the polar axis (θ), measured counterclockwise in radians.
• Non-uniqueness: the same point can be represented infinitely many ways by adding multiples of 2π to θ or using negative r-values.
• Grid structure: polar graph paper uses concentric circles (constant r) and radial lines (constant θ) instead of horizontal and vertical lines.
• Common confusion: converting between systems requires attention to quadrant—tan⁻¹(y/x) alone doesn't determine θ; you must check which quadrant the point lies in.
• When polar is better: equations for circles, spheres, and phenomena with radial symmetry are often simpler in polar form than Cartesian.

📐 Coordinate system basics

📐 How polar coordinates work

Polar coordinates of a point P in the plane are (r, θ), where:

• |r| is the distance from P to the pole
• θ is the angle measured counterclockwise from the polar axis to the ray through P from the pole

• The system starts with a single point (the pole, equivalent to the origin) and a single ray (the polar axis).
• Instead of "left/right and up/down," you think "how far and in which direction."
• Angles are always measured in radians.
• Example: The point (2, π/2) means "move 2 units from the pole at angle π/2 radians."

🗺️ Polar graph paper structure

Feature	Polar	Cartesian
Grid lines	Concentric circles and radial lines	Horizontal and vertical lines
Constant r	Points on a circle	Not a simple shape
Constant θ	Points on a ray from pole	Not a simple shape
Constant x or y	Not a simple shape	Vertical or horizontal line

• Circles centered at the pole have equation r = k (constant).
• Lines through the pole have equation θ = k (constant).
• This grid helps locate points by distance and angle.

🔄 Non-uniqueness of polar coordinates

🔄 Multiple representations of the same point

Three ways the same point can have different polar coordinates:

1. Adding multiples of 2π: (r, θ) is the same as (r, θ + 2kπ) for any integer k, because angles repeat every full rotation.
2. Negative r-values: (r, θ) is the same as (−r, θ + π), meaning you move in the opposite direction.
3. The pole itself: (0, θ) represents the pole for any value of θ.

Example: The point (1, π/2) can also be written as (1, 5π/2) or (1, −3π/2) or (−1, 3π/2).

⚠️ Don't confuse: positive vs negative r

• With positive r at angle θ, you move in the direction of θ.
• With negative r at angle θ, you move in the opposite direction (as if the angle were θ + π).
• Example: (2, π/4) is in the first quadrant, but (−2, π/4) is in the third quadrant—the same location as (2, 5π/4).

🗺️ Regions and inequalities

🗺️ Describing regions with polar inequalities

Simple regions in polar coordinates:

• Annular ring (ring shape): a ≤ r ≤ b describes all points between distance a and distance b from the pole, with no restriction on angle.
• Circular sector (pie wedge): a ≤ θ ≤ b with r ≤ c describes a wedge-shaped region.
• Example: 1 ≤ r < 3 is a ring; 0 ≤ r ≤ 2, π/6 < θ < π/3 is a wedge.

These regions are simpler to describe in polar form than in Cartesian form when they have radial symmetry.

🔀 Converting between coordinate systems

🔀 Polar to Cartesian conversion

Formulas:

• x = r cos(θ)

• y = r sin(θ)

• These come from the right triangle formed by the point, with r as hypotenuse.

• Example: To convert (√2, 3π/4) to Cartesian:

  • x = √2 cos(3π/4) = √2 · (−1/√2) = −1
  • y = √2 sin(3π/4) = √2 · (1/√2) = 1
  • Result: (−1, 1)

🔀 Cartesian to polar conversion

Formulas:

• r = √(x² + y²)
• tan(θ) = y/x

Critical step: After computing tan(θ), you must choose θ in the correct quadrant.

• Computing tan⁻¹(y/x) alone is not enough—there are always two angles between 0 and 2π with the same tangent.
• You must check which quadrant the original point (x, y) lies in.
• Example: For (−1/2, −√3/2), tan(θ) = √3, but the point is in the third quadrant, so θ = 4π/3, not π/3.

⚠️ Don't confuse: tan⁻¹ doesn't give the full answer

The excerpt emphasizes: "it is not enough to compute tan⁻¹(y/x); we must choose the angle in the same quadrant as the given point."

🔄 Converting equations

🔄 Cartesian to polar equation conversion

Strategy: Replace x with r cos(θ) and y with r sin(θ).

• Example: 2x + 3y = 6 becomes 2r cos(θ) + 3r sin(θ) = 6, which simplifies to r = 6/(2 cos(θ) + 3 sin(θ)).
• Not all equations become simpler—lines are better suited to Cartesian coordinates.
• Example: x² + y² = 4 becomes r² = 4, which simplifies to r = 2 (much simpler!).

🔄 Polar to Cartesian equation conversion

Strategies to try:

• Replace r cos(θ) with x
• Replace r sin(θ) with y
• Replace r² with x² + y²
• Replace tan(θ) with y/x

Example: r = 3 cos(θ) converts as follows:

• Multiply both sides by r: r² = 3r cos(θ)
• Replace: x² + y² = 3x
• Complete the square: (x − 3/2)² + y² = 9/4
• This is a circle with center (3/2, 0) and radius 3/2.

🎯 When polar coordinates shine

The excerpt notes: "Polar coordinates are useful for studying objects or phenomena that have radial symmetry, such as circles, spheres, and cylinders, or the central forces (those that act equally in all directions), such as gravity and electric charge."

• Equations for these objects are often simpler in polar form.
• Circles centered at the pole have the simple equation r = constant.
• Objects with rotational symmetry around a point are naturally described in polar terms.

🧭 Overview

🧠 One-sentence thesis

Polar graphs are traced by sweeping counterclockwise around the pole, where the r-value at each angle θ determines the distance from the origin, producing distinctive families of curves like roses, limaçons, and spirals that differ fundamentally from Cartesian graphing.

📌 Key points (3–5)

• How polar graphing differs from Cartesian: In Cartesian coordinates you move horizontally then vertically; in polar coordinates the dependent variable r gives a distance from the pole in direction θ, not a height.
• Standard polar curve families: Circles, roses (with petals), limaçons (including cardioids), lemniscates, and spirals each have characteristic equations and shapes.
• Common confusion—tracing vs. existence: A polar graph may be traced multiple times as θ ranges from 0 to 2π, and negative r-values plot points in the opposite direction, so the same geometric curve can correspond to different θ intervals.
• Finding intersections requires extra care: Solving r = f(θ) and r = g(θ) simultaneously may miss the pole, because r = 0 can occur at different θ values in each equation—always check the pole separately.
• Guidepoints simplify sketching: For standard curves, locate key points (petal tips, quadrantal angles, maximum/minimum r) to sketch the graph quickly by hand.

📐 How polar graphing works

📐 The sweeping-radius mental model

When graphing an equation in polar coordinates, we think of sweeping around the pole in the counterclockwise direction, and at each angle θ the r-value tells us how far the graph is from the pole.

• In Cartesian coordinates, you build a graph left-to-right with "height" at each x.
• In polar coordinates, you rotate counterclockwise from the positive x-axis; at each angle θ, the equation gives a radius r.
• The graph is the set of all points (r, θ) that satisfy the equation.
• Example: For r = 2 sin(θ), as θ increases from 0 to π/2, r increases from 0 to 2, so a radial line sweeping through the first quadrant grows longer, tracing out part of a circle.

🔄 Negative r-values and multiple traces

• When r is negative, the point is plotted in the opposite direction from angle θ.
• Example: The point (−1, π/6) is the same as (1, 7π/6)—both represent the same location, but with different (r, θ) coordinates.
• A single geometric curve may be traced more than once as θ ranges from 0 to 2π.
• Don't confuse: "The graph exists only for certain θ" vs. "the graph is traced multiple times"—the same curve can be drawn repeatedly with different parameter values.

🌸 Standard polar curve families

🌸 Roses: r = a sin(nθ) or r = a cos(nθ)

A rose with petal length a. If n is odd, the rose has n petals; if n is even, it has 2n petals.

• Petal length: The parameter a determines how far each petal extends from the pole.
• Number of petals: Depends on n—odd n gives n petals, even n gives 2n petals.
• Sketching strategy: Find the tips of the petals by solving for when r = a (maximum value). For r = 3 sin(2θ), set 3 = 3 sin(2θ), so sin(2θ) = 1, giving 2θ = π/2, hence θ = π/4. The four petal tips are evenly spaced π/2 apart: π/4, 3π/4, 5π/4, 7π/4.
• Example: r = 2 cos(3θ) is a rose with 3 petals of length 2.

🫘 Limaçons: r = a ± b sin(θ) or r = a ± b cos(θ)

• Shape depends on a vs. b:
  • If a > b: the limaçon has a dent (like a lima bean).
  • If a < b: the limaçon has an inner loop.
  • If a = b: the limaçon is a cardioid (heart-shaped).
• Symmetry: Limaçons with cosine are symmetric about the x-axis; those with sine are symmetric about the y-axis.
• Sketching strategy: Evaluate r at the four quadrantal angles (θ = 0, π/2, π, 3π/2) as guidepoints, then connect smoothly.
• Example: r = 3 + 2 cos(θ) has a = 3, b = 2, so a > b → dent-shaped limaçon. At θ = 0, r = 5; at θ = π/2, r = 3; at θ = π, r = 1; at θ = 3π/2, r = 3.

⭕ Circles in polar form

Equation	Description
r = k (constant)	Circle centered at the pole with radius k
r = 2a sin(θ)	Circle with center (0, a) on the y-axis, radius |a|
r = 2a cos(θ)	Circle with center (a, 0) on the x-axis, radius |a|

• Verification by conversion: The equation r = 2 sin(θ) can be rewritten as r² = 2r sin(θ), then x² + y² = 2y, which completes the square to (x − 0)² + (y − 1)² = 1, confirming a circle with center (0, 1) and radius 1.

🦋 Lemniscates: r² = a² cos(2θ) or r² = a² sin(2θ)

• Figure-eight or "infinity symbol" shape.
• The equation involves r², so you must solve for r = ±√[a² cos(2θ)] to graph.
• Why some θ are excluded: When cos(2θ) or sin(2θ) is negative, r² would be negative, which is impossible—so no points exist for those angles.
• Example: r² = 4 cos(2θ) has no points for π/4 ≤ θ ≤ 3π/4 because cos(2θ) is negative there.

🌀 Spirals

• Archimedean spiral: r = aθ. As θ increases, r increases at a constant rate, winding outward steadily.
• Logarithmic spiral: r = e^(aθ). As θ increases, r grows exponentially, winding outward faster and faster.

🔍 Finding intersection points

🔍 Solve the system algebraically

• To find intersections of r = f(θ) and r = g(θ), set f(θ) = g(θ) and solve for θ.
• Then substitute back to find the corresponding r-values.
• Example: For r = 2 + 2 sin(θ) and r = 2 + 2 cos(θ), set 2 + 2 sin(θ) = 2 + 2 cos(θ), so sin(θ) = cos(θ), giving tan(θ) = 1. Solutions: θ = π/4 and θ = 5π/4. Evaluating r gives intersection points (2 + √2, π/4) and (2 − √2, 5π/4).

⚠️ Always check the pole separately

Solving a system of equations r = f(θ) and r = g(θ) will not always reveal an intersection at the pole, because r may be equal to zero for different values of θ in the two equations.

• The pole has infinitely many representations: (0, θ) for any θ.
• Two curves may both pass through the pole but at different θ values.
• How to check: Set r = 0 in each equation separately and solve for θ. If both equations have solutions (even different ones), the pole is an intersection point.
• Example: For r = 2 + 2 sin(θ) and r = 2 + 2 cos(θ), setting r = 0 gives sin(θ) = −1 (so θ = 3π/2) for the first and cos(θ) = −1 (so θ = π) for the second. Both represent the pole, confirming it is a third intersection point.

🎨 Sketching techniques

🎨 Use guidepoints for standard curves

• For roses: Find petal tips by solving for maximum r, then space them evenly (angle between petals = 2π divided by number of petals).
• For limaçons and cardioids: Evaluate r at θ = 0, π/2, π, 3π/2, plot these four points, and connect smoothly.
• For circles: Identify the center and radius from the equation form, or convert to Cartesian to verify.

🔄 Connect points in order of increasing θ

• The order matters: polar graphs are traced as θ increases, not drawn all at once.
• Connecting points out of order will produce the wrong shape.
• Example: For r = 2 sin(3θ), the first loop is traced for 0 ≤ θ ≤ π/3, the second loop for π/3 ≤ θ ≤ 2π/3, and the third for 2π/3 ≤ θ ≤ π. Each loop corresponds to a specific θ interval.

🖩 Use a calculator to explore, then sketch by hand

• Set the calculator to Polar mode and enter appropriate window settings (θ min, θ max, θ step, and x/y ranges).
• Observe how the graph is traced as θ increases.
• Once you recognize the curve type, sketch it by hand using guidepoints and symmetry.

🧭 Overview

🧠 One-sentence thesis

Complex numbers extend the real number system to include solutions to equations like x² + 1 = 0, enabling every polynomial to have exactly n zeros and providing a complete algebraic framework for solving all polynomial equations.

📌 Key points (3–5)

• Why complex numbers exist: Real numbers cannot solve equations like x² − 2x + 2 = 0 because they require square roots of negative numbers; complex numbers fill this gap.
• Structure of complex numbers: Every complex number has the form z = a + bi, where a is the real part, b is the imaginary part, and i² = −1.
• Arithmetic operations: Complex numbers can be added, subtracted, multiplied, and divided using algebraic rules; division requires multiplying by the complex conjugate.
• Common confusion—real vs. complex plane: In the Cartesian plane, one point (x, y) represents two separate real numbers; in the complex plane, one point represents a single complex number z = a + bi.
• Fundamental theorem of algebra: Every polynomial of degree n ≥ 1 has exactly n complex zeros (counting multiplicities), and nonreal zeros of polynomials with real coefficients always occur in conjugate pairs.

🔢 The imaginary unit and imaginary numbers

🔢 Defining i

Imaginary unit: i is defined by i² = −1, or equivalently i = √(−1).

• The letter i is not a variable; it is a constant representing a specific number.
• This definition allows us to work with square roots of negative numbers.

🧮 Imaginary numbers

Imaginary number: For any real number a > 0, √(−a) = √(−1) √a = i√a.

• Any number that is the product of i and a real number is called an imaginary number.
• Examples: 3i, (7/8)i, −38i, i√5.
• Example: √(−25) = √(−1) √25 = i · 5 = 5i; and 2√(−3) = 2√(−1) √3 = 2i√3.
• Just as positive numbers have two real square roots, every negative number has two imaginary square roots (e.g., the square roots of −9 are 3i and −3i).

⚠️ Caution about products

• The identity √(ab) = √a √b fails when both a and b are negative.
• Example: If a = b = −2, then √(ab) = √(4) = 2, but √a · √b = √(−2) · √(−2) = i√2 · i√2 = i² · 2 = −2, so √(ab) ≠ √a √b.
• Best practice: Always write square roots of negative numbers as imaginary numbers before performing calculations.

🧩 Structure of complex numbers

🧩 Definition and parts

Complex number: A complex number can be written in the form z = a + bi, where a and b are real numbers.

• Real part: a is called the real part of z.
• Imaginary part: b is called the imaginary part of z (note: b itself is a real number; the imaginary part is the coefficient of i).
• Examples: 3 − 5i, 2 + √7 i, 4 − (i/3), 6i, −9.
• All real numbers are also complex numbers (with imaginary part b = 0).
• A pure imaginary number has real part a = 0.

🔍 Equality of complex numbers

Equality: Two complex numbers z₁ = a + bi and z₂ = c + di are equal if and only if a = c and b = d.

• The real and imaginary parts cannot be combined.
• Two complex numbers are equal only when both their real parts match and their imaginary parts match.

🔗 Complex conjugate

Complex conjugate: For any complex number z = a + bi, the number z̄ = a − bi is called the complex conjugate of z.

• The conjugate has the same real part but the opposite imaginary part.
• Key property: The product of a nonzero complex number and its conjugate is always a positive real number: z z̄ = (a + bi)(a − bi) = a² + b² (since i² = −1).
• Example: If z = 7 − 5i, then z̄ = 7 + 5i, and z z̄ = 49 − 25i² = 49 + 25 = 74.

➕ Arithmetic operations

➕ Addition and subtraction

Sum: z₁ + z₂ = (a + bi) + (c + di) = (a + c) + (b + d)i
Difference: z₁ − z₂ = (a + bi) − (c + di) = (a − c) + (b − d)i

• Combine real parts separately and imaginary parts separately.
• Example: (4 + 5i) + (2 − 3i) = (4 + 2) + (5 − 3)i = 6 + 2i.
• Example: (8 − 6i) − (5 + 2i) = (8 − 5) + (−6 − 2)i = 3 − 8i.

✖️ Multiplication

Product: z₁z₂ = (a + bi)(c + di) = (ac − bd) + (ad + bc)i

• Use the distributive law as if multiplying binomials, then replace i² with −1.
• Example: (2 + 3i)(3 − 5i) = 6 − 10i + 9i − 15i² = 6 − 10i + 9i + 15 = 21 − i.
• Example: (3i)(4i) = 12i² = 12(−1) = −12.

➗ Division

Quotient: (z₁/z₂) = (a + bi)/(c + di) = [(a + bi)/(c + di)] · [(c − di)/(c − di)] = [(ac + bd)/(c² + d²)] + [(bc − ad)/(c² + d²)]i

• Multiply numerator and denominator by the complex conjugate of the denominator.
• This rationalizes the denominator, making it a positive real number.
• Example: (5)/(2 + 3i) · (2 − 3i)/(2 − 3i) = (10 − 15i)/13 = (10/13) − (15/13)i.
• Example: (2 + 3i)/(4 − 2i) = [(2 + 3i)(4 + 2i)]/[(4 − 2i)(4 + 2i)] = (8 + 4i + 12i + 6i²)/(16 + 4) = (2 + 16i)/20 = (1/10) + (4/5)i.

📐 The complex plane

📐 Graphing complex numbers

Complex plane: A plane in which the horizontal axis represents real numbers (real axis) and the vertical axis represents pure imaginary numbers (imaginary axis).

• To plot z = a + bi, move a units horizontally and b units vertically from the origin.
• Example: To plot z = 2 + 3i, move 2 units right and 3 units up; to plot z̄ = 2 − 3i, move 2 units right and 3 units down.
• Don't confuse: In the Cartesian plane, the point (2, −6) represents two separate real numbers x = 2 and y = −6; in the complex plane, the point (2, −6) represents the single complex number z = 2 − 6i.

📏 Modulus

Modulus: The modulus of a complex number z = a + bi is |z| = √(a² + b²).

• The modulus is the distance from the origin to the point (a, b) in the complex plane (analogous to absolute value for real numbers).
• Computed using the Pythagorean theorem.
• Example: |2 + 3i| = √(4 + 9) = √13.
• Example: The equation |z| = 4 describes a circle of radius 4 centered at the origin.

🔺 Vector representation and addition

• A complex number z = a + bi can be viewed as a vector (arrow) from the origin to the point (a, b).
• Vector addition: The sum z₁ + z₂ corresponds to the diagonal of the parallelogram formed by the vectors z₁ and z₂ as adjacent sides.
• Example: To add z₁ = 5 − i and z₂ = −7 + 4i, draw vectors to (5, −1) and (−7, 4), form a parallelogram, and the diagonal ends at z₁ + z₂ = −2 + 3i.
• Subtraction: z₁ − z₂ = z₁ + (−z₂), where −z₂ is a vector with the same length as z₂ but pointing in the opposite direction.

🌳 Zeros of polynomials

🌳 Evaluating polynomials at complex numbers

• Because we can add, subtract, and multiply complex numbers, we can evaluate any polynomial at a complex input.
• Example: For f(x) = x² − 2x + 2, evaluate f(1 + i):
  f(1 + i) = (1 + i)² − 2(1 + i) + 2 = 1 + 2i + i² − 2 − 2i + 2 = 1 + 2i − 1 − 2 − 2i + 2 = 0.
  So 1 + i is a zero of the polynomial.

🎯 Quadratic polynomials and conjugate pairs

• The zeros of ax² + bx + c are given by the quadratic formula: x = [−b ± √(b² − 4ac)]/(2a).
• If the discriminant D = b² − 4ac is negative, the two solutions are complex conjugates.
• Example: For x² − 4x + 5 = 0, the solutions are [4 ± √(−4)]/2 = (4 ± 2i)/2 = 2 ± i.
• Key fact: If z = a + bi (b ≠ 0) is one zero of a quadratic polynomial with real coefficients, then z̄ = a − bi is the other zero.

🏗️ Constructing quadratic polynomials

Quadratic with complex zeros: If z = a + bi (b ≠ 0) is a zero of a quadratic polynomial p(x) with real coefficients and lead coefficient 1, then p(x) = (x − z)(x − z̄) = x² − (z + z̄)x + zz̄.

• Both (z + z̄) and zz̄ are real numbers, so the polynomial has real coefficients.
• Example: To find a quadratic with one zero z = 7 − 5i:
  z̄ = 7 + 5i, z + z̄ = 14, zz̄ = 49 + 25 = 74.
  So p(x) = x² − 14x + 74.

🏆 Fundamental theorem of algebra

Fundamental theorem of algebra: Let p(x) be a polynomial of degree n ≥ 1. Then p(x) has exactly n complex zeros, counting multiplicities.

• A zero "of multiplicity k" means the factor (x − a) occurs k times.
• Example: p(x) = (x − 5)² has a zero of multiplicity 2 at x = 5.
• Every polynomial of degree n can be factored as the product of n linear terms (over the complex numbers).
• Example: Although x⁴ + 4 has no real zeros, it has four complex zeros: 1 + i, −1 + i, −1 − i, 1 − i, and factors as [x − (1 + i)][x − (−1 + i)][x − (−1 − i)][x − (1 − i)].

🔄 Conjugate pairs for higher-degree polynomials

• For any polynomial with real coefficients, nonreal zeros always occur in complex conjugate pairs.
• Example: To find a fourth-degree polynomial with zeros 3i and 2 + i:
  The other two zeros must be −3i and 2 − i.
  Factored form: (x − 3i)(x + 3i)[x − (2 + i)][x − (2 − i)] = (x² + 9)(x² − 4x + 5) = x⁴ − 4x³ + 14x² − 36x + 45.

📊 Summary table: Operations on complex numbers

Operation	Formula	Key idea
Addition	(a + bi) + (c + di) = (a + c) + (b + d)i	Combine real and imaginary parts separately
Subtraction	(a + bi) − (c + di) = (a − c) + (b − d)i	Combine real and imaginary parts separately
Multiplication	(a + bi)(c + di) = (ac − bd) + (ad + bc)i	Use distributive law; replace i² with −1
Division	(a + bi)/(c + di) = [(ac + bd)/(c² + d²)] + [(bc − ad)/(c² + d²)]i	Multiply by conjugate of denominator
Conjugate product	(a + bi)(a − bi) = a² + b²	Always a positive real number

🧭 Overview

🧠 One-sentence thesis

The polar form of complex numbers provides a powerful way to visualize and compute products, quotients, powers, and roots by representing complex numbers in terms of their distance from the origin and angle from the real axis.

📌 Key points (3–5)

• Polar form representation: Any complex number z = a + bi can be written as z = r(cos(θ) + i sin(θ)), where r is the modulus (distance from origin) and θ is the argument (angle from real axis).
• Multiplication by i means rotation: Multiplying a complex number by i rotates its vector representation by 90° counterclockwise around the origin.
• Products and quotients are simpler: In polar form, multiply complex numbers by multiplying their moduli and adding their arguments; divide by dividing moduli and subtracting arguments.
• Powers via De Moivre's Theorem: To raise a complex number to the nth power, raise the modulus to the nth power and multiply the argument by n.
• Finding roots: A complex number has n distinct nth roots, evenly spaced around a circle, found by taking the nth root of the modulus and dividing the argument by n (plus adding multiples of 2π/n).

🔄 Converting between forms

🔄 From Cartesian to polar

Polar form: z = r(cos(θ) + i sin(θ)), where r = √(a² + b²) and θ is defined by a = r cos(θ), b = r sin(θ), 0 ≤ θ ≤ 2π

• Start with z = a + bi (Cartesian form)
• Calculate the modulus r = √(a² + b²) (distance from origin)
• Find the argument θ using tan(θ) = b/a, being careful about which quadrant the complex number is in
• Example: For z = √3 + i, we get r = 2 and θ = π/6, so z = 2(cos(π/6) + i sin(π/6))

🔄 From polar to Cartesian

• Given z = r(cos(θ) + i sin(θ))
• Evaluate the trigonometric functions: a = r cos(θ), b = r sin(θ)
• Write as z = a + bi
• Example: z = 2(cos(π/6) + i sin(π/6)) = 2(√3/2 + i·1/2) = √3 + i

🔄 Terminology

• Argument: the angle θ in polar form
• Modulus: the distance r from the origin (also called "length")
• Don't confuse: The argument is not unique—you can add any multiple of 2π and get the same complex number

🔁 Multiplication by i as rotation

🔁 Geometric interpretation

• When you multiply a complex number by i, its graph rotates 90° counterclockwise around the origin
• Example from the text: z = 2 + 3i becomes iz = 2i - 3 (which is -3 + 2i)
• The vector representing iz is perpendicular to the vector representing z
• This suggests that multiplication by complex numbers involves rotation—polar coordinates are well-suited for rotation

🔁 Why polar form helps

• Polar coordinates use angles to specify location
• Rotation is naturally described by adding to an angle
• This makes polar form ideal for understanding complex multiplication

✖️ Products and quotients in polar form

✖️ Product formula

Product in polar form: If z₁ = r(cos(α) + i sin(α)) and z₂ = R(cos(β) + i sin(β)), then z₁z₂ = rR(cos(α + β) + i sin(α + β))

• To multiply: multiply the moduli and add the arguments
• Example: z = 2(cos(π/6) + i sin(π/6)) and w = 2(cos(π/3) + i sin(π/3))
  • Product: zw = 4(cos(π/2) + i sin(π/2)) = 4i
• Geometric meaning: multiplying by w rotates z by the argument of w (in this case, π/3 or 60°)

✖️ Quotient formula

Quotient in polar form: If z₁ = r(cos(α) + i sin(α)) and z₂ = R(cos(β) + i sin(β)), then z₁/z₂ = (r/R)(cos(α - β) + i sin(α - β))

• To divide: divide the moduli and subtract the arguments
• Example: For the same z and w above, z/w = 1(cos(-π/6) + i sin(-π/6)) = √3/2 - (1/2)i

🔢 Powers using De Moivre's Theorem

🔢 The theorem

De Moivre's Theorem: If z = r(cos(α) + i sin(α)) is a complex number in polar form, and n is a positive integer, then zⁿ = rⁿ(cos(nα) + i sin(nα))

• To raise to a power: raise the modulus to that power and multiply the argument by that power
• Example: (√3 + i)⁴
  • First convert: z = 2(cos(π/6) + i sin(π/6))
  • Apply theorem: z⁴ = 2⁴(cos(4·π/6) + i sin(4·π/6)) = 16(cos(2π/3) + i sin(2π/3))
  • Convert back: 16(-1/2 + i√3/2) = -8 + 8i√3

🔢 Why it works

• Raising to a power is repeated multiplication
• Each multiplication adds the argument again
• So multiplying n times means adding the argument n times, which is the same as multiplying by n

🌿 Roots of complex numbers

🌿 Finding nth roots

Roots of a complex number: A complex number z = r(cos(α) + i sin(α)) has n complex nth roots, given by zₖ = r^(1/n)(cos((α + 2πk)/n) + i sin((α + 2πk)/n)) for k = 0, 1, 2, ..., n-1

• Every number has n distinct nth roots
• To find them: take the nth root of the modulus, divide the argument by n, then add multiples of 2π/n
• The roots are evenly spaced around a circle of radius r^(1/n)

🌿 Example: square roots

• To find square roots of w = -8 + 8i√3:
  • Convert to polar: w = 16(cos(2π/3) + i sin(2π/3))
  • Can also write as: w = 16(cos(8π/3) + i sin(8π/3)) (adding 2π)
  • First root: z₁ = 4(cos(π/3) + i sin(π/3)) = 2 + 2i√3
  • Second root: z₂ = 4(cos(4π/3) + i sin(4π/3)) = -2 - 2i√3

🌿 Example: cube roots of 8

• One polar form: z = 8(cos 0 + i sin 0)
• Other forms: z = 8(cos 2π + i sin 2π) and z = 8(cos 4π + i sin 4π)
• Three cube roots:
  • z₁ = 2(cos 0 + i sin 0) = 2
  • z₂ = 2(cos(2π/3) + i sin(2π/3)) = -1 + i√3
  • z₃ = 2(cos(4π/3) + i sin(4π/3)) = -1 - i√3
• These are evenly spaced around a circle of radius 2

🌿 Geometric pattern

• The n nth roots are evenly distributed around a circle
• They are separated by angles of 2π/n
• Don't confuse: Adding more multiples of 2π doesn't give new roots—you cycle back to the ones you already found