College Trigonometry

🧭 Overview

🧠 One-sentence thesis

Angles can be measured in degrees or radians, and radian measure naturally connects angles to real numbers through the unit circle, enabling applications in circular motion and beyond.

📌 Key points (3–5)

• Two measurement systems: Degrees (360° per revolution) and radians (2π per revolution) both quantify rotation between two rays sharing a common vertex.
• Radian measure advantage: On the unit circle, an angle's radian measure equals the arc length it subtends, directly identifying angles with real numbers.
• Oriented angles: Direction matters—positive angles rotate counter-clockwise, negative angles rotate clockwise, and coterminal angles differ by full revolutions (360° or 2π).
• Common confusion: Radian measure is dimensionless (a ratio of lengths), so "π/6 radians" is really just the number π/6; don't treat radians as physical units like degrees.
• Circular motion application: For constant angular velocity ω on a circle of radius r, linear velocity v = rω connects rotational speed to tangential speed.

📐 Basic angle concepts

📐 What is an angle

Ray: A half-line with one endpoint (the initial point) extending infinitely in one direction.

Angle: Formed when two rays share a common initial point (the vertex).

• An angle measures the amount of rotation separating its two rays.
• The same diagram can represent multiple angles depending on which rotation path you measure.
• Example: Two rays can form both a small angle and a large angle going "the other way around."

🏷️ Labeling and types

• Angles are labeled with Greek letters: α (alpha), β (beta), γ (gamma), θ (theta).
• Special cases:
  • Straight angle: The two rays point in opposite directions (180°).
  • Right angle: One-quarter revolution (90°), marked with a small square symbol.
  • Acute angle: Strictly between 0° and 90°.
  • Obtuse angle: Strictly between 90° and 180°.

🌐 Degree measure

🌐 How degrees work

Degree measure: One complete revolution equals 360°, and partial rotations are measured proportionally.

• One degree (1°) = 1/360 of a full revolution.
• Half revolution = 180°, quarter revolution = 90°.
• Any angle's measure can be determined by knowing what fraction of a full revolution it represents.
• Example: An angle representing 2/3 of a revolution measures (2/3)(360°) = 240°.

🔢 Subdividing degrees

Two systems exist for angles smaller than 1°:

System	Description	Example
Decimal degrees	Use decimals (e.g., 30.5°)	30.5° is halfway between 30° and 31°
DMS (Degree-Minute-Second)	1° = 60′, 1′ = 60″, so 1° = 3600″	42.125° = 42°7′30″

Converting decimal to DMS:

• Separate whole degrees from the decimal part.
• Multiply the decimal part by 60 to get minutes (separate whole and decimal again).
• Multiply remaining decimal by 60 to get seconds.

Converting DMS to decimal:

• Convert minutes to degrees by dividing by 60.
• Convert seconds to degrees by dividing by 3600.
• Add all parts together.

🔗 Angle relationships

Complementary angles: Two acute angles whose measures sum to 90°.

Supplementary angles: Two angles (either both right, or one acute and one obtuse) whose measures sum to 180°.

Don't confuse: These relationships depend on the sum, not the individual angle types.

↻ Oriented angles and standard position

↻ Direction matters

Oriented angle: An angle where rotation direction is specified—positive for counter-clockwise, negative for clockwise.

• Angles can represent more than one full revolution.
• Example: 450° = one full revolution (360°) plus an additional 90°.
• The distinction between "the angle" and "its measure" is often blurred in practice (e.g., "α = 42°").

📍 Standard position

Standard position: An angle whose vertex is at the origin and whose initial side lies on the positive x-axis.

Classification by terminal side location:

Terminal side location	Name
In Quadrant I, II, III, or IV	"Quadrant [I/II/III/IV] angle"
On a coordinate axis	"Quadrantal angle"

🔄 Coterminal angles

Coterminal angles: Angles in standard position that share the same terminal side.

• Coterminal angles differ by multiples of 360°.
• Formula: If α and β are coterminal, then β = α + 360°·k for some integer k.
• Example: 120° and −240° are coterminal because −240° = 120° − 360°.
• Every angle has infinitely many coterminal angles (one for each integer k).

🔵 Radian measure

🔵 Definition and motivation

π (pi): The ratio of any circle's circumference to its diameter; π = C/d.

• Since diameter = 2·radius, we have 2π = C/r (circumference-to-radius ratio is constant).
• For an arc of length s on a circle of radius r, the ratio s/r is also constant.

Radian measure: The measure of a central angle θ equals s/r, where s is the arc length the angle subtends and r is the circle's radius.

Intuitive meaning: An angle of 1 radian means the arc length equals the radius; 2 radians means arc length = 2r, etc.

🔵 Key radian values

• One full revolution: 2πr/r = 2π radians
• Half revolution: π radians
• Quarter revolution: π/2 radians
• Radians are dimensionless (length ÷ length), so they're "pure numbers"

🔵 Oriented radians

• Positive radian measure: counter-clockwise rotation
• Negative radian measure: clockwise rotation
• Coterminal angles in radians: β = α + 2πk for integer k
• Complementary angles: α + β = π/2
• Supplementary angles: α + β = π

🔄 Converting between degrees and radians

Conversion factors:

• Degrees to radians: multiply by (π radians)/(180°)
• Radians to degrees: multiply by (180°)/(π radians)

Degrees	Radians	Calculation
60°	π/3	60°·(π/180°) = π/3
−150°	−5π/6	(−5π/6)·(180°/π) = −150°
1 radian	≈57.3°	180°/π ≈ 57.2958°

Don't confuse: The negative sign carries through in both systems—it always indicates clockwise rotation.

🎯 The unit circle connection

🎯 Identifying angles with real numbers

On the unit circle (x² + y² = 1):

• For an angle θ in standard position, let s = arc length from (1,0) to the terminal point.
• Since radius = 1, the radian measure equals the arc length: θ = s/1 = s.
• This directly identifies each real number t with an angle of t radians.

Wrapping the number line:

• For t > 0: wrap the interval [0, t] counter-clockwise around the unit circle starting at (1, 0).
• For t < 0: wrap the interval [t, 0] clockwise around the unit circle.
• For t = 0: the point is (1, 0).

Example: The real number 3π/4 corresponds to an angle of 3π/4 radians, which is 3/8 of a revolution counter-clockwise from (1, 0).

🎯 Large and negative values

• For |t| > 2π, the arc wraps around the circle multiple times.
• Example: t = 117 ≈ 18.62 revolutions, so after 18 full loops, about 0.62 of a revolution remains.
• For negative t, count clockwise revolutions.

Don't confuse: The arc length and angle measure are the same number on the unit circle, but this only works because radius = 1.

🎡 Circular motion applications

🎡 Velocity concepts

Setup: An object moves on a circular path of radius r from point P to point Q in time t.

Displacement s: The arc length traveled (positive for counter-clockwise, negative for clockwise).

Average velocity v̄: The average rate of change of position = s/t (units: length/time).

• The formula θ = s/r still holds with signed values.
• Speed = |v̄| (the magnitude, ignoring direction).

🎡 Angular velocity

Average angular velocity ω̄: The average rate of change of angle = θ/t (units: radians/time).

Since s = rθ, we have: v̄ = s/t = rθ/t = r·(θ/t) = r·ω̄

When angular velocity is constant:

• Average velocities equal instantaneous velocities (v̄ = v, ω̄ = ω).
• v is called linear velocity (or just "velocity").
• ω is called angular velocity.

🎡 The fundamental formula

v = rω: For circular motion with constant angular velocity ω on a circle of radius r, the linear velocity is v = rω.

Unit reconciliation:

• Left side: length/time
• Right side: length · (radians/time) = (length · radians)/time
• Since radians are dimensionless, this reduces to length/time ✓

Physical interpretation: Points farther from the center (larger r) must travel faster (larger v) to maintain the same angular frequency ω, because they cover more distance per revolution.

🎡 Frequency and period

Quantity	Symbol	Definition	Units
Ordinary frequency	f	Revolutions per unit time	1/time
Angular frequency	ω	Radians per unit time; ω = 2πf	radians/time
Period	T	Time for one complete cycle; T = 1/f	time

Example: Earth rotates once per 24 hours, so f = 1/(24 hours) and ω = 2π/(24 hours) = π/12 radians per hour. At latitude with radius 2960 miles, linear velocity v = 2960·(π/12) ≈ 775 miles/hour.

Don't confuse: Frequency measures "how often," period measures "how long for one cycle"—they're reciprocals.

🧭 Overview

🧠 One-sentence thesis

The cosine and sine functions assign to each angle a unique position on the Unit Circle—cosine as the x-coordinate and sine as the y-coordinate—enabling us to describe circular motion, solve triangles, and establish fundamental trigonometric relationships.

📌 Key points (3–5)

• Core definitions: For an angle θ in standard position, cos(θ) is the x-coordinate and sin(θ) is the y-coordinate of the point where the terminal side intersects the Unit Circle.
• The Pythagorean Identity: cos²(θ) + sin²(θ) = 1 always holds because the point lies on the Unit Circle (x² + y² = 1).
• Reference angles: For non-quadrantal angles, the reference angle α (the acute angle to the x-axis) lets you find cos(θ) and sin(θ) using symmetry: cos(θ) = ±cos(α) and sin(θ) = ±sin(α), with signs determined by the quadrant.
• Common confusion: Angles can be coterminal (differ by integer multiples of 2π) and have the same cosine and sine values; also, the same angle can be written in multiple equivalent forms (e.g., −π/6 and 11π/6 differ by 2π).
• Beyond the unit circle: For circles of radius r, the coordinates become (r cos(θ), r sin(θ)), which describes circular motion and enables right-triangle trigonometry.

📍 Defining cosine and sine on the Unit Circle

📍 The basic definitions

Cosine of θ: the x-coordinate of the point P where the terminal side of θ (in standard position) intersects the Unit Circle.

Sine of θ: the y-coordinate of that same point P.

• These definitions make cosine and sine functions: each angle θ maps to exactly one cos(θ) value and one sin(θ) value.
• The point P is written as (cos(θ), sin(θ)).

🔢 Finding values for specific angles

Quadrantal angles (terminal side on an axis) are straightforward:

• θ = 270° → terminal side on negative y-axis → P = (0, −1) → cos(270°) = 0, sin(270°) = −1.
• θ = −π (half clockwise revolution) → terminal side on negative x-axis → P = (−1, 0) → cos(−π) = −1, sin(−π) = 0.

Non-quadrantal angles require geometry:

• For θ = 45°, drop a perpendicular to form a 45°–45°–90° triangle; using x² + y² = 1 and y = x, solve to get cos(45°) = sin(45°) = √2/2.
• For θ = π/6 (30°), a 30°–60°–90° triangle gives sin(π/6) = 1/2 and cos(π/6) = √3/2.
• For θ = 60°, the same triangle approach yields cos(60°) = 1/2 and sin(60°) = √3/2.

Don't confuse: The process uses the Unit Circle equation x² + y² = 1 and special-triangle geometry; the values are exact, not approximations.

🔗 The Pythagorean Identity

🔗 Where it comes from

• Since P(cos(θ), sin(θ)) lies on the Unit Circle, substituting x = cos(θ) and y = sin(θ) into x² + y² = 1 gives:
  • cos²(θ) + sin²(θ) = 1
• Notation: (cos(θ))² is written cos²(θ); (sin(θ))² is written sin²(θ).

The Pythagorean Identity: For any angle θ, cos²(θ) + sin²(θ) = 1.

🔗 How to use it

• If you know one of cos(θ) or sin(θ), you can find the other up to a ± sign.
• The quadrant where θ's terminal side lies removes the ambiguity:
  • Example: If θ is Quadrant II and sin(θ) = 3/5, then cos²(θ) = 1 − 9/25 = 16/25, so cos(θ) = ±4/5. Since x-coordinates are negative in Quadrant II, cos(θ) = −4/5.
  • Example: If sin(θ) = 1, then cos²(θ) = 0, so cos(θ) = 0 (no ambiguity).

Don't confuse: The ± comes from algebra; the quadrant determines which sign is correct.

🪞 Symmetry and reference angles

🪞 What a reference angle is

Reference angle α: For a non-quadrantal angle θ, α is the acute angle between the terminal side of θ and the x-axis.

• Quadrant I or IV: α is measured from the positive x-axis.
• Quadrant II or III: α is measured from the negative x-axis.

🪞 The Reference Angle Theorem

Theorem: If α is the reference angle for θ, then cos(θ) = ±cos(α) and sin(θ) = ±sin(α), where the signs depend on the quadrant.

• The Unit Circle has symmetry (x-axis, y-axis, origin), so points symmetric to (cos(α), sin(α)) determine the values for other quadrants.
• Example: For θ = 5π/6, the terminal side is π/6 short of π (half revolution). The reference angle is α = π/6. Since θ is Quadrant II, cos(5π/6) = −cos(π/6) = −√3/2 and sin(5π/6) = sin(π/6) = 1/2.

📋 Essential values to memorize

θ (degrees)	θ (radians)	cos(θ)	sin(θ)
0°	0	1	0
30°	π/6	√3/2	1/2
45°	π/4	√2/2	√2/2
60°	π/3	1/2	√3/2
90°	π/2	0	1

• Use these with the Reference Angle Theorem to find cosine and sine for angles in any quadrant.

🪞 Examples using reference angles

• θ = 225°: Quadrant III, overshoots 180° by 45°, so α = 45°. Both cos and sin are negative: cos(225°) = −√2/2, sin(225°) = −√2/2.
• θ = 11π/6: Quadrant IV, just shy of 2π, so α = 2π − 11π/6 = π/6. Cosine positive, sine negative: cos(11π/6) = √3/2, sin(11π/6) = −1/2.
• θ = −5π/4: Clockwise 5π/4 lands in Quadrant II, reference angle π/4. Cosine negative, sine positive: cos(−5π/4) = −√2/2, sin(−5π/4) = √2/2.
• θ = 7π/3: More than 2π, so subtract 2π to get coterminal angle π/3 (Quadrant I): cos(7π/3) = 1/2, sin(7π/3) = √3/2.

Tip: In radian measure, the denominator hints at the reference angle—denominator 6 → π/6, denominator 4 → π/4, denominator 3 → π/3.

🧮 Solving basic trigonometric equations

🧮 Finding all angles that satisfy an equation

• Solutions typically involve infinitely many angles because cosine and sine repeat every 2π (coterminal angles).
• General form: find one solution in each relevant quadrant, then add integer multiples of 2π.

🧮 Examples

cos(θ) = 1/2:

• The x-coordinate is 1/2 → Quadrant I or IV, reference angle π/3.
• Quadrant I: θ = π/3 + 2πk for integers k.
• Quadrant IV: θ = 5π/3 + 2πk for integers k.

sin(θ) = −1/2:

• The y-coordinate is −1/2 → Quadrant III or IV, reference angle π/6.
• Quadrant III: θ = 7π/6 + 2πk.
• Quadrant IV: θ = 11π/6 + 2πk.

cos(θ) = 0:

• Terminal side on the y-axis → θ = π/2 or 3π/2.
• Can write as θ = π/2 + πk for integers k (captures both).

Don't confuse: Different-looking families can represent the same set of angles (e.g., −π/6 + 2πk and 11π/6 + 2πk are equivalent because 11π/6 = −π/6 + 2π).

🌐 Beyond the Unit Circle: circles of radius r

🌐 Coordinates on a circle of radius r

Theorem: If Q(x, y) is the point on the terminal side of θ (in standard position) that lies on the circle x² + y² = r², then:

• x = r cos(θ) and y = r sin(θ)
• cos(θ) = x/r and sin(θ) = y/r

• This generalizes the Unit Circle (where r = 1).
• Proof uses similar triangles: the triangle formed by Q on the larger circle is similar to the triangle formed by the corresponding point on the Unit Circle.

🌐 Application: circular motion

For an object moving in a circular path of radius r with constant angular velocity ω, starting at (r, 0) at time t = 0:

• The angle swept out is θ = ωt.
• The position at time t is:
  • x = r cos(ωt)
  • y = r sin(ωt)
• ω > 0 indicates counter-clockwise; ω < 0 indicates clockwise.

Example: Lakeland Community College at latitude 41.628°N, radius 2960 miles, angular velocity π/12 radians per hour:

• x = 2960 cos(πt/12), y = 2960 sin(πt/12), where x and y are in miles and t is in hours.

🌐 Application: right-triangle trigonometry

Theorem: For an acute angle θ in a right triangle with adjacent side a, opposite side b, and hypotenuse c:

• cos(θ) = a/c (adjacent over hypotenuse)
• sin(θ) = b/c (opposite over hypotenuse)

• This follows from placing the triangle in standard position and using the circle of radius c.
• Example: A 30° angle with adjacent side 7. Find the hypotenuse c and opposite side b.
  • cos(30°) = 7/c → c = 7/cos(30°) = 7/(√3/2) = 14√3/3.
  • sin(30°) = b/c → b = c sin(30°) = (14√3/3)(1/2) = 7√3/3.

Don't confuse: This is the same cosine and sine from the Unit Circle, just applied to a different radius.

🔢 Cosine and sine as functions of real numbers

🔢 Extending to real numbers

• Identify each real number t with the angle θ = t radians.
• Define cos(t) = cos(θ) and sin(t) = sin(θ).
• Equivalently: wrap an oriented arc of length t (starting at (1, 0)) around the Unit Circle to reach P(cos(t), sin(t)).

🔢 Domain and range

Domain: Both f(t) = cos(t) and g(t) = sin(t) are defined for all real numbers t, so domain is (−∞, ∞).

Range: Since cos(t) and sin(t) are coordinates on the Unit Circle, both take all values between −1 and 1, so range is [−1, 1].

Don't confuse: Whether you think of t as a real number or as an angle in radians, the process and results are identical; the distinction is often blurred in practice.

---

Summary: The Unit Circle provides a geometric foundation for cosine and sine, linking angles to coordinates. The Pythagorean Identity, reference angles, and symmetry make it possible to find exact values efficiently. Extending to circles of any radius connects trigonometry to motion, geometry, and real-world applications.

🧭 Overview

🧠 One-sentence thesis

Cosine and sine, defined via coordinates on the Unit Circle, are the foundation of six circular functions that can be expressed through fundamental algebraic identities.

📌 Key points (3–5)

• What "circular functions" means: cosine and sine earn this name because they are defined using coordinate values of points on the Unit Circle.
• Why cosine and sine are foundational: the excerpt states they are "just" the beginning—implying other functions build on them.
• Common confusion: the excerpt ends mid-sentence, so the full set of six functions and their identities are not yet presented in this fragment.
• Context from prior sections: Section 10.2 already established cos(θ) and sin(θ) definitions; this section (10.3) extends that framework.

📭 Content limitation notice

📭 Incomplete excerpt

The provided source text for Section 10.3 contains only the opening sentence:

"In section 10.2, we defined cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions. It turns out that cosine and sine are just..."

• The sentence breaks off before explaining what the six circular functions are.
• No fundamental identities are presented.
• No definitions, theorems, examples, or exercises from Section 10.3 are included.

🔍 What we know from context

• From the title: this section will cover six circular functions (not just cosine and sine) and fundamental identities relating them.
• From Section 10.2: cosine and sine are defined as the x- and y-coordinates of a point on the Unit Circle where the terminal side of angle θ intersects the circle.
• Implication: the remaining four circular functions are likely defined in terms of cosine and sine (or their ratios), but the excerpt does not provide this information.

🚧 Missing body sections

Because the excerpt contains no substantive content from Section 10.3, the following topics cannot be covered:

• Definitions of the four additional circular functions (likely tangent, cotangent, secant, cosecant).
• Fundamental identities (e.g., reciprocal identities, quotient identities, Pythagorean-related identities).
• Examples demonstrating how to use these identities.
• Comparisons or common confusions between the six functions.

To study Section 10.3 effectively, you will need the complete text.

🧭 Overview

🧠 One-sentence thesis

The Unit Circle definitions of cosine and sine enable us to find exact trigonometric values through reference angles, symmetry, and the Pythagorean Identity, which together form the foundation for solving trigonometric equations and modeling circular motion.

📌 Key points (3–5)

• Cosine and sine as coordinates: For an angle θ in standard position, cos(θ) and sin(θ) are the x- and y-coordinates of the point where the terminal side intersects the Unit Circle.
• The Pythagorean Identity: cos²(θ) + sin²(θ) = 1 always holds because the point lies on the Unit Circle (x² + y² = 1).
• Reference angles simplify computation: Any non-quadrantal angle's cosine and sine equal ±cos(α) and ±sin(α) of its reference angle α, with signs determined by the quadrant.
• Common confusion—quadrantal vs non-quadrantal: Quadrantal angles (0°, 90°, 180°, 270°) have terminal sides on axes and are easy to evaluate; non-quadrantal angles require reference angles or special triangles.
• Extension to real numbers and motion: Identifying real number t with θ = t radians allows cosine and sine to model circular motion: x = r cos(ωt), y = r sin(ωt).

📐 Defining cosine and sine on the Unit Circle

📍 What the definitions mean

Cosine of θ: the x-coordinate of the point P where the terminal side of θ (in standard position) intersects the Unit Circle.

Sine of θ: the y-coordinate of that same point P.

• These are functions: each angle θ has exactly one cos(θ) and one sin(θ).
• The point P = (cos(θ), sin(θ)) always lies on the circle x² + y² = 1.
• Example: For θ = 270°, the terminal side lies on the negative y-axis, so P = (0, −1), giving cos(270°) = 0 and sin(270°) = −1.

🔺 Finding values with special triangles

When the terminal side does not lie on an axis, drop a perpendicular to the x-axis to form a right triangle.

• 45°–45°–90° triangle: Both legs equal, so for θ = 45°, cos(45°) = sin(45°) = √2/2.
• 30°–60°–90° triangle: For θ = 30° (π/6), sin(30°) = 1/2 and cos(30°) = √3/2; for θ = 60° (π/3), sin(60°) = √3/2 and cos(60°) = 1/2.
• Use the Pythagorean Theorem on the triangle and the fact that P lies on the Unit Circle to solve for unknown coordinates.

🔗 The Pythagorean Identity

🧮 Where it comes from

Since P(cos(θ), sin(θ)) lies on the Unit Circle x² + y² = 1, substituting x = cos(θ) and y = sin(θ) yields:

Pythagorean Identity: cos²(θ) + sin²(θ) = 1 for any angle θ.

• The notation cos²(θ) means (cos(θ))²; similarly for sin²(θ).
• This identity is always true, regardless of θ.

🔍 Using the identity to find missing values

If one of cos(θ) or sin(θ) is known, the identity determines the other up to a ± sign.

• Removing the ambiguity: Knowing which quadrant the terminal side of θ lies in tells you the sign.
  • Quadrant I: both positive
  • Quadrant II: cos negative, sin positive
  • Quadrant III: both negative
  • Quadrant IV: cos positive, sin negative
• Example: If sin(θ) = 3/5 and θ is in Quadrant II, then cos²(θ) = 1 − 9/25 = 16/25, so cos(θ) = ±4/5. Since Quadrant II has negative x-coordinates, cos(θ) = −4/5.

🪞 Reference angles and symmetry

🔄 What a reference angle is

Reference angle α: the acute angle between the terminal side of θ and the x-axis.

• For Quadrant I or IV angles, α is measured from the positive x-axis.
• For Quadrant II or III angles, α is measured from the negative x-axis.
• α is always acute (between 0 and 90°, or 0 and π/2 radians).

🎯 Reference Angle Theorem

cos(θ) = ±cos(α) and sin(θ) = ±sin(α), where the signs depend on the quadrant of θ.

• The Unit Circle has symmetry with respect to both axes and the origin, so any point P on the circle has a symmetric counterpart Q that determines the reference angle.
• Example: For θ = 5π/6 (150°), the terminal side is π/6 short of π, so α = π/6. Since θ is in Quadrant II, cos(5π/6) = −cos(π/6) = −√3/2 and sin(5π/6) = sin(π/6) = 1/2.

🗂️ Essential values to memorize

θ (degrees)	θ (radians)	cos(θ)	sin(θ)
0°	0	1	0
30°	π/6	√3/2	1/2
45°	π/4	√2/2	√2/2
60°	π/3	1/2	√3/2
90°	π/2	0	1

• Use these values plus the Reference Angle Theorem to find cosine and sine for any common angle.
• Don't confuse: Reduced fractions with denominator 6 have reference angle π/6; denominator 4 → π/4; denominator 3 → π/3.

🔢 Solving basic trigonometric equations

🎲 Finding all angles that satisfy an equation

Trigonometric equations typically have infinitely many solutions because cosine and sine repeat every full rotation (2π radians or 360°).

• Example: Solve cos(θ) = 1/2.
  • The terminal side intersects the Unit Circle at x = 1/2, which occurs in Quadrant I and Quadrant IV.
  • Reference angle is π/3.
  • Quadrant I solution: θ = π/3 + 2πk for integers k.
  • Quadrant IV solution: θ = 5π/3 + 2πk for integers k.
• The "+ 2πk" captures all coterminal angles (angles that differ by full rotations).

⚠️ Common confusion—different-looking families can be the same

• The Quadrant IV solution to sin(θ) = −1/2 can be written θ = 11π/6 + 2πk or θ = −π/6 + 2πk.
• These represent the same set of angles because −π/6 and 11π/6 are coterminal.
• When in doubt, write out a few values from each family to verify they match.

🌍 Beyond the Unit Circle—circles of radius r

📏 Extending to any circle centered at the origin

For a circle x² + y² = r², if Q(x, y) is the point on the terminal side of θ that lies on this circle:

x = r cos(θ) and y = r sin(θ)

Conversely:

cos(θ) = x/r = x/√(x² + y²) and sin(θ) = y/r = y/√(x² + y²)

• When r = 1 (the Unit Circle), these reduce to the original definitions.
• Example: If the terminal side of θ contains the point Q(4, −2), then r = √(16 + 4) = 2√5, so cos(θ) = 4/(2√5) = 2√5/5 and sin(θ) = −2/(2√5) = −√5/5.

🌐 Application—modeling circular motion

An object moving in a circular path of radius r with constant angular velocity ω has position at time t:

x = r cos(ωt) and y = r sin(ωt)

• Assumes the object starts at (r, 0) when t = 0.
• ω > 0: counter-clockwise; ω < 0: clockwise.
• Example: The Earth's rotation at a given latitude can be modeled this way to find the position of a location over time.

📐 Right triangle trigonometry

🔺 Relating angles to side lengths

For an acute angle θ in a right triangle:

cos(θ) = (adjacent side)/hypotenuse and sin(θ) = (opposite side)/hypotenuse

• Adjacent side: the leg next to θ (not the hypotenuse).
• Opposite side: the leg across from θ.
• Hypotenuse: the side opposite the right angle.
• This follows from placing the triangle in standard position on the Unit Circle and scaling by the hypotenuse length.

🧮 Solving right triangles

Given one acute angle and one side, use the definitions to find the other sides.

• Example: If θ = 30° and the adjacent side is 7, then cos(30°) = 7/c, so c = 7/cos(30°) = 7/(√3/2) = 14√3/3.
• Then use sin(30°) = b/c to find the opposite side b, or use the Pythagorean Theorem.
• Don't confuse: You can use either trigonometry or the Pythagorean Theorem to find a missing side; both methods should agree.

🔢 Cosine and sine as functions of real numbers

🔄 Identifying real numbers with radian angles

A real number t corresponds to the angle θ = t radians, so:

cos(t) = cos(θ) and sin(t) = sin(θ)

• Equivalently, t is the length of an oriented arc on the Unit Circle starting at (1, 0) and ending at P(cos(t), sin(t)).
• This allows us to treat cosine and sine as functions of real numbers, not just angles.

📊 Domain and range

Function	Domain	Range
f(t) = cos(t)	(−∞, ∞)	[−1, 1]
g(t) = sin(t)	(−∞, ∞)	[−1, 1]

• Why the domain is all reals: Every real number t corresponds to some angle in radian measure.
• Why the range is [−1, 1]: Cosine and sine are coordinates on the Unit Circle, which has radius 1.
• Solving equations like sin(t) = −1/2 uses the same method as solving sin(θ) = −1/2; only the variable name changes.

🧭 Overview

🧠 One-sentence thesis

The cosine and sine functions assign to every angle a unique position on the Unit Circle—cosine as the x-coordinate and sine as the y-coordinate—and this geometric definition unlocks a powerful framework for describing circular motion, solving triangles, and understanding fundamental trigonometric relationships.

📌 Key points (3–5)

• Core definitions: cosine of θ is the x-coordinate and sine of θ is the y-coordinate of the point where the terminal side of θ intersects the Unit Circle.
• The Pythagorean Identity: cos²(θ) + sin²(θ) = 1 for any angle, derived from the Unit Circle equation x² + y² = 1.
• Reference angles: for non-quadrantal angles, the acute angle between the terminal side and the x-axis determines cosine and sine values up to sign; the quadrant determines the sign.
• Common confusion: distinguishing which quadrant an angle lies in is essential—the same reference angle yields different signs in different quadrants.
• Beyond the unit circle: the definitions extend to circles of any radius r and enable modeling circular motion and solving right triangles.

📐 Defining cosine and sine on the Unit Circle

📍 What cosine and sine represent

Cosine of θ: the x-coordinate of the point P where the terminal side of θ (in standard position) intersects the Unit Circle.

Sine of θ: the y-coordinate of that same point P.

• For each angle θ, there is exactly one point P on the Unit Circle, so cosine and sine are well-defined functions.
• Notation: P = (cos(θ), sin(θ)).
• Example: For θ = 270°, the terminal side lies on the negative y-axis, so P = (0, −1), giving cos(270°) = 0 and sin(270°) = −1.

🔺 Finding values using triangles

When the terminal side does not lie on a coordinate axis, drop a perpendicular from P to the x-axis to form a right triangle.

• For θ = 45°, the triangle is a 45°–45°–90° triangle; using geometry and the Unit Circle equation x² + y² = 1, we find cos(45°) = sin(45°) = √2/2.
• For θ = π/6 (30°), a 30°–60°–90° triangle yields sin(π/6) = 1/2 and cos(π/6) = √3/2.
• For θ = 60° (π/3), the same triangle gives cos(60°) = 1/2 and sin(60°) = √3/2.

Don't confuse: the process uses the Unit Circle constraint (x² + y² = 1) and geometric properties of special triangles; the coordinates must satisfy both.

🔑 The Pythagorean Identity

🔑 Derivation and meaning

Since P(cos(θ), sin(θ)) lies on the Unit Circle x² + y² = 1, substituting gives:

Pythagorean Identity: cos²(θ) + sin²(θ) = 1 for any angle θ.

• The notation cos²(θ) means (cos(θ))²; similarly for sin²(θ).
• This identity is always true, regardless of θ.
• It connects the Pythagorean Theorem to the circle equation.

🧮 Using the identity to find missing values

If one of cos(θ) or sin(θ) is known, the identity determines the other up to a ± sign.

• The quadrant where the terminal side lies removes the ambiguity.
• Example: If θ is in Quadrant II and sin(θ) = 3/5, then cos²(θ) = 1 − 9/25 = 16/25, so cos(θ) = ±4/5. Since x-coordinates are negative in Quadrant II, cos(θ) = −4/5.
• Example: If sin(θ) = 1, then cos²(θ) = 0, so cos(θ) = 0 (no ambiguity here).

Don't confuse: the ± from algebra vs. the actual sign determined by the quadrant.

🪞 Symmetry and reference angles

🪞 What is a reference angle?

Reference angle α: for a non-quadrantal angle θ, the acute angle between the terminal side of θ and the x-axis.

• If θ is in Quadrant I or IV, α is measured from the positive x-axis.
• If θ is in Quadrant II or III, α is measured from the negative x-axis.
• The Unit Circle's symmetry (with respect to both axes and the origin) means points related by reflection share the same reference angle.

🎯 Reference Angle Theorem

Theorem: If α is the reference angle for θ, then cos(θ) = ±cos(α) and sin(θ) = ±sin(α), where the sign depends on the quadrant of θ.

• This reduces all cosine and sine calculations to a small set of known values (0°, 30°, 45°, 60°, 90°).
• Example: θ = 5π/6 lies in Quadrant II, π/6 short of π. Its reference angle is π/6. Since Quadrant II has negative x and positive y, cos(5π/6) = −√3/2 and sin(5π/6) = 1/2.
• Example: θ = 225° is in Quadrant III with reference angle 45°. Both coordinates are negative, so cos(225°) = sin(225°) = −√2/2.

Table of essential values (must memorize):

θ (degrees)	θ (radians)	cos(θ)	sin(θ)
0°	0	1	0
30°	π/6	√3/2	1/2
45°	π/4	√2/2	√2/2
60°	π/3	1/2	√3/2
90°	π/2	0	1

🔄 Coterminal angles

Angles that differ by full rotations (multiples of 2π or 360°) share the same terminal side, hence the same cosine and sine.

• Example: θ = 7π/3 = 2π + π/3, so cos(7π/3) = cos(π/3) = 1/2 and sin(7π/3) = sin(π/3) = √3/2.

🔢 Solving basic trigonometric equations

🔢 Finding all angles with a given cosine or sine

Because cosine and sine are periodic (repeat every 2π), equations like cos(θ) = 1/2 have infinitely many solutions.

• Identify which quadrants satisfy the equation (by sign).
• Find one solution in each quadrant using reference angles.
• Add integer multiples of 2π to capture all coterminal solutions.

Example: cos(θ) = 1/2

• The x-coordinate is 1/2 in Quadrants I and IV.
• Reference angle is π/3.
• Quadrant I: θ = π/3 + 2πk for integers k.
• Quadrant IV: θ = 5π/3 + 2πk for integers k.

Example: sin(θ) = −1/2

• The y-coordinate is −1/2 in Quadrants III and IV.
• Reference angle is π/6.
• Quadrant III: θ = 7π/6 + 2πk.
• Quadrant IV: θ = 11π/6 + 2πk.

Example: cos(θ) = 0

• Terminal side lies on the y-axis.
• θ = π/2 + πk for integers k (covers both π/2 and 3π/2 plus all coterminals).

Don't confuse: different-looking families can represent the same set of angles (e.g., θ = −π/6 + 2πk is the same as θ = 11π/6 + 2πk).

🌐 Beyond the Unit Circle

🌐 Circles of radius r

For a point Q(x, y) on a circle of radius r centered at the origin, where the terminal side of θ intersects:

Theorem: x = r cos(θ) and y = r sin(θ).

Equivalently:

• cos(θ) = x/r = x/√(x² + y²)

• sin(θ) = y/r = y/√(x² + y²)

• When r = 1 (Unit Circle), this reduces to the original definitions.

• Example: If the terminal side contains Q(4, −2), then r = √(16 + 4) = 2√5, so cos(θ) = 4/(2√5) = 2√5/5 and sin(θ) = −2/(2√5) = −√5/5.

🌍 Application: circular motion

An object moving in a circular path of radius r with constant angular velocity ω has position at time t given by:

Equations of motion: x = r cos(ωt) and y = r sin(ωt), assuming the object starts at (r, 0) when t = 0.

• ω > 0 indicates counter-clockwise motion; ω < 0 indicates clockwise.
• Example: Lakeland Community College at latitude 41.628°N rotates with the Earth. Using r ≈ 2960 miles and ω = π/12 radians per hour, the motion is x = 2960 cos(πt/12) and y = 2960 sin(πt/12).

📐 Right triangle trigonometry

For an acute angle θ in a right triangle:

Theorem: cos(θ) = (adjacent side)/(hypotenuse) and sin(θ) = (opposite side)/(hypotenuse).

• This follows from placing the triangle in standard position and applying the circle-based definitions.
• Example: In a right triangle with a 30° angle, adjacent side 7, we find the hypotenuse c using cos(30°) = 7/c, so c = 7/(√3/2) = 14√3/3. Then the opposite side b = c sin(30°) = (14√3/3)(1/2) = 7√3/3.

Don't confuse: "adjacent" and "opposite" are relative to the angle θ, not the triangle as a whole.

🔢 Cosine and sine as functions of real numbers

🔢 Domain and range

By identifying a real number t with the angle θ = t radians, we extend cosine and sine to functions of real numbers.

Domain: both f(t) = cos(t) and g(t) = sin(t) have domain (−∞, ∞).

Range: both have range [−1, 1], since coordinates on the Unit Circle are bounded.

• Solving sin(t) = −1/2 for real t is done exactly as for angles in radians: t = 7π/6 + 2πk or t = 11π/6 + 2πk.
• The distinction between "t as a real number" and "θ = t radians" is often blurred in practice; properties developed for angles in radians apply equally to real-number inputs.

Don't confuse: the input can be thought of as either a real number or an angle in radians—the mathematics is the same.

🧭 Overview

🧠 One-sentence thesis

The cosine and sine functions assign coordinates on the Unit Circle to every angle, providing a foundation for describing circular motion and solving right triangles through reference angles and the Pythagorean Identity.

📌 Key points (3–5)

• Core definitions: Cosine is the x-coordinate and sine is the y-coordinate of the point where an angle's terminal side intersects the Unit Circle.
• The Pythagorean Identity: For any angle θ, cos²(θ) + sin²(θ) = 1, derived from the Unit Circle equation x² + y² = 1.
• Reference angles simplify computation: Any non-quadrantal angle's cosine and sine can be found using its acute reference angle plus appropriate signs based on quadrant.
• Common confusion: The same angle can have multiple representations (e.g., 11π/6 and −π/6 + 2πk describe the same set of angles); solutions to trig equations typically involve infinitely many coterminal angles.
• Extension beyond the unit circle: The definitions extend to circles of any radius r and to real numbers (not just angles), enabling equations of circular motion.

📐 Defining cosine and sine on the Unit Circle

📍 Basic definition from coordinates

Cosine of θ: the x-coordinate of the point P where the terminal side of angle θ (in standard position) intersects the Unit Circle.

Sine of θ: the y-coordinate of that same point P.

• Notation: If P is the intersection point, then P = (cos(θ), sin(θ)).
• These are functions: each angle θ produces exactly one cosine value and one sine value.
• Example: For θ = 270°, the terminal side lies on the negative y-axis, so the point is (0, −1), giving cos(270°) = 0 and sin(270°) = −1.

🔺 Finding values using geometry

For quadrantal angles (0°, 90°, 180°, 270°):

• The terminal side lies on a coordinate axis.
• Simply read off the coordinates: (1, 0), (0, 1), (−1, 0), (0, −1).

For non-quadrantal angles (like 45°, 30°, 60°):

• Drop a perpendicular from point P to the x-axis, forming a right triangle.
• Use special triangle properties (45-45-90 or 30-60-90) plus the Unit Circle equation x² + y² = 1.
• Example: For θ = 45°, the triangle has equal legs, so x = y. Substituting into x² + y² = 1 gives 2x² = 1, so x = √2/2. Hence cos(45°) = sin(45°) = √2/2.

📊 Essential values to memorize

θ (degrees)	θ (radians)	cos(θ)	sin(θ)
0°	0	1	0
30°	π/6	√3/2	1/2
45°	π/4	√2/2	√2/2
60°	π/3	1/2	√3/2
90°	π/2	0	1

• The excerpt emphasizes these must be memorized.
• Quadrantal angles (multiples of 90°) are also essential.

🔗 The Pythagorean Identity

🧮 Derivation and statement

• Since P(x, y) = (cos(θ), sin(θ)) lies on the Unit Circle, it satisfies x² + y² = 1.
• Substituting x = cos(θ) and y = sin(θ) gives: (cos(θ))² + (sin(θ))² = 1.
• Convention (which the excerpt calls "unfortunate"): write this as cos²(θ) + sin²(θ) = 1.

Theorem 10.1 (Pythagorean Identity): For any angle θ, cos²(θ) + sin²(θ) = 1.

🔄 Using the identity to find missing values

• If one of cos(θ) or sin(θ) is known, the identity determines the other up to a ± sign.
• The quadrant where θ's terminal side lies removes the ambiguity.
• Example: If θ is in Quadrant II and sin(θ) = 3/5, then cos²(θ) = 1 − 9/25 = 16/25, so cos(θ) = ±4/5. Since x-coordinates are negative in Quadrant II, cos(θ) = −4/5.
• Don't confuse: the identity always holds, but the sign depends on the quadrant.

🪞 Reference angles and symmetry

🔍 What is a reference angle?

Reference angle α for a non-quadrantal angle θ: the acute angle between the terminal side of θ and the x-axis.

• If θ is in Quadrant I or IV, α is measured from the positive x-axis.
• If θ is in Quadrant II or III, α is measured from the negative x-axis.
• The Unit Circle's symmetry (with respect to both axes and the origin) means every angle's cosine and sine relate to its reference angle's values.

📐 Reference Angle Theorem

Theorem 10.2: If α is the reference angle for θ, then cos(θ) = ±cos(α) and sin(θ) = ±sin(α), where the ± signs depend on which quadrant θ's terminal side lies in.

How to apply:

1. Plot θ in standard position.
2. Identify the quadrant.
3. Find the reference angle α (the acute angle to the nearest part of the x-axis).
4. Use the memorized values for α.
5. Attach the correct signs based on the quadrant (x-coordinate sign for cosine, y-coordinate sign for sine).

Example: For θ = 225°, the terminal side is in Quadrant III, 45° past the negative x-axis, so α = 45°. Both cosine and sine are negative in Quadrant III, so cos(225°) = −cos(45°) = −√2/2 and sin(225°) = −sin(45°) = −√2/2.

🔢 Spotting reference angles in radian measure

• Reduced fractions with denominator 6 → reference angle π/6.
• Denominator 4 → reference angle π/4.
• Denominator 3 → reference angle π/3.
• Example: 11π/6 is in Quadrant IV; reference angle is 2π − 11π/6 = π/6.

🔄 Solving basic trigonometric equations

🎯 Finding all angles that satisfy an equation

• Trigonometric equations typically have infinitely many solutions because angles repeat every full rotation (360° or 2π radians).
• Strategy: find one solution in each relevant quadrant, then add integer multiples of 2π (for coterminal angles).

Example: Solve cos(θ) = 1/2.

• The terminal side intersects the Unit Circle at x = 1/2.
• This occurs in Quadrant I (θ = π/3) and Quadrant IV (θ = 5π/3).
• All solutions: θ = π/3 + 2πk or θ = 5π/3 + 2πk for any integer k.

⚠️ Common confusion: equivalent solution families

• Different-looking answers can represent the same set of angles.
• Example: For sin(θ) = −1/2 in Quadrant IV, one answer is θ = 11π/6 + 2πk; another is θ = −π/6 + 2πk. Both describe the same infinite list of coterminal angles.
• The excerpt advises: "When in doubt, write it out!" — list a few values from each family to verify they match.

🌐 Beyond the Unit Circle

📏 Extending to circles of radius r

• For a circle x² + y² = r² centered at the origin, if Q(x, y) is on the terminal side of θ:
  • x = r cos(θ)
  • y = r sin(θ)
• Conversely: cos(θ) = x/r and sin(θ) = y/r, where r = √(x² + y²).

Theorem 10.3: If Q(x, y) lies on the terminal side of angle θ (in standard position) on the circle x² + y² = r², then x = r cos(θ), y = r sin(θ), and cos(θ) = x/√(x² + y²), sin(θ) = y/√(x² + y²).

• This reduces to the Unit Circle definitions when r = 1.

🔁 Equations of circular motion

• An object moving in a circular path of radius r with constant angular velocity ω has position at time t given by:
  • x = r cos(ωt)
  • y = r sin(ωt)
• Assumes the object starts at (r, 0) when t = 0.
• ω > 0 indicates counter-clockwise motion; ω < 0 indicates clockwise.
• Example: If r = 2960 miles and ω = π/12 radians per hour, then x = 2960 cos(πt/12) and y = 2960 sin(πt/12), with x and y in miles and t in hours.

📐 Right triangle trigonometry

🔺 Relating angles to side lengths

• For an acute angle θ in a right triangle:
  • Adjacent side: the leg next to θ (length a).
  • Opposite side: the leg across from θ (length b).
  • Hypotenuse: the side opposite the right angle (length c).

Theorem 10.4: For an acute angle θ in a right triangle, cos(θ) = (adjacent)/(hypotenuse) = a/c and sin(θ) = (opposite)/(hypotenuse) = b/c.

• This follows from placing the triangle in standard position on a circle of radius c and applying Theorem 10.3.
• Example: In a right triangle with a 30° angle, adjacent side 7, find the hypotenuse c. Using cos(30°) = 7/c and cos(30°) = √3/2, we get c = 7/(√3/2) = 14√3/3.

🧭 Don't confuse: Unit Circle vs. right triangle perspectives

• The Unit Circle definition works for any angle (including obtuse, negative, or angles greater than 360°).
• The right triangle definition only applies to acute angles (0° < θ < 90°).
• Both perspectives give the same values for acute angles, but the Unit Circle approach is more general.

🔢 Cosine and sine as functions of real numbers

🔄 Extending to real number inputs

• Any real number t can be identified with the angle θ = t radians.
• Define cos(t) = cos(θ) and sin(t) = sin(θ).
• This means expressions like cos(π) or sin(2) can be evaluated by treating the input as an angle in radians or as a real number—the choice is equivalent.

📊 Domain and range

Theorem 10.5:

• The functions f(t) = cos(t) and g(t) = sin(t) both have domain (−∞, ∞).
• Both have range [−1, 1].

Why:

• Every real number t corresponds to some angle, so cosine and sine are defined for all real numbers.
• Since cos(t) and sin(t) represent coordinates on the Unit Circle, they take on all values between −1 and 1, inclusive.

🔁 Solving equations with real number variables

• Equations like sin(t) = −1/2 are solved the same way as angle equations.
• The solution uses reference angles/arcs: t = 7π/6 + 2πk or t = 11π/6 + 2πk for integers k.
• The distinction between "t as a real number" and "θ = t radians as an angle" is often blurred in practice; the excerpt notes this is mostly cosmetic.

🧭 Overview

🧠 One-sentence thesis

The cosine and sine functions assign to every angle a unique position on the Unit Circle, enabling us to describe circular motion, solve right triangles, and establish fundamental relationships through the Pythagorean Identity.

📌 Key points (3–5)

• What cosine and sine are: For any angle θ in standard position, cos(θ) is the x-coordinate and sin(θ) is the y-coordinate of the point where the terminal side intersects the Unit Circle.
• The Pythagorean Identity: Because points lie on the Unit Circle (x² + y² = 1), we always have cos²(θ) + sin²(θ) = 1, which helps find one value when the other is known.
• Reference angles simplify computation: The acute angle between the terminal side and the x-axis (the reference angle) lets us use symmetry to find cosine and sine values in any quadrant.
• Common confusion—quadrantal vs non-quadrantal angles: Quadrantal angles (0°, 90°, 180°, 270°) lie on axes and have easy values (0, ±1); non-quadrantal angles require geometry or reference angles.
• Extension to circles of any radius: The formulas x = r cos(θ) and y = r sin(θ) describe positions on circles of radius r, enabling equations of circular motion.

📍 Defining cosine and sine on the Unit Circle

📍 The basic definitions

Cosine of θ: the x-coordinate of the point P where the terminal side of θ (in standard position) intersects the Unit Circle.
Sine of θ: the y-coordinate of that same point P.

• In symbols: P = (cos(θ), sin(θ)).
• These are functions: each angle θ produces exactly one cos(θ) and one sin(θ).
• Example: For θ = 270°, the terminal side lies on the negative y-axis, so the point is (0, −1), giving cos(270°) = 0 and sin(270°) = 0.

🔺 Finding values with right triangles

When the terminal side does not lie on an axis, drop a perpendicular from P to the x-axis to form a right triangle.

• For θ = 45°, the triangle is 45°–45°–90°, so both legs are equal; using x² + y² = 1 and y = x gives cos(45°) = sin(45°) = √2/2.
• For θ = 30° (or π/6), a 30°–60°–90° triangle yields sin(π/6) = 1/2 and cos(π/6) = √3/2.
• For θ = 60° (or π/3), the same triangle gives cos(60°) = 1/2 and sin(60°) = √3/2.
• Don't confuse: the process is the same, but the specific triangle (45°–45°–90° vs 30°–60°–90°) determines which values you get.

🔗 The Pythagorean Identity

🔗 Why it's always true

Because P(x, y) = (cos(θ), sin(θ)) lies on the Unit Circle, it satisfies x² + y² = 1.

• Substituting x = cos(θ) and y = sin(θ) gives (cos(θ))² + (sin(θ))² = 1.
• Convention (unfortunate but standard): write this as cos²(θ) + sin²(θ) = 1.
• This is called the Pythagorean Identity because it derives from the Pythagorean Theorem via the circle equation.

🔗 Using the identity to find missing values

If you know one of cos(θ) or sin(θ), the identity gives the other up to a ± sign; the quadrant removes the ambiguity.

• Example: If sin(θ) = 3/5 and θ is in Quadrant II, then cos²(θ) = 1 − 9/25 = 16/25, so cos(θ) = ±4/5. Since x-coordinates are negative in Quadrant II, cos(θ) = −4/5.
• Example: If sin(θ) = 1, then cos²(θ) = 0, so cos(θ) = 0 (no ambiguity).

🪞 Reference angles and symmetry

🪞 What a reference angle is

Reference angle α for θ: the acute angle between the terminal side of θ and the x-axis.

• If θ is in Quadrant I or IV, α is measured from the positive x-axis.
• If θ is in Quadrant II or III, α is measured from the negative x-axis.
• The Unit Circle's symmetry (across x-axis, y-axis, and origin) means the point P for θ is a reflection of the point Q for α.

🪞 The Reference Angle Theorem

Theorem: cos(θ) = ± cos(α) and sin(θ) = ± sin(α), where the signs depend on which quadrant θ is in.

• Quadrant I: both positive.
• Quadrant II: cos negative, sin positive.
• Quadrant III: both negative.
• Quadrant IV: cos positive, sin negative.
• Example: θ = 225° is in Quadrant III with reference angle α = 225° − 180° = 45°, so cos(225°) = −cos(45°) = −√2/2 and sin(225°) = −sin(45°) = −√2/2.

📋 Essential values to memorize

The excerpt provides a table of common angles (in degrees and radians) with their cosine and sine values:

θ (degrees)	θ (radians)	cos(θ)	sin(θ)
0°	0	1	0
30°	π/6	√3/2	1/2
45°	π/4	√2/2	√2/2
60°	π/3	1/2	√3/2
90°	π/2	0	1

• These values, combined with the Reference Angle Theorem, generate all the important points on the Unit Circle.
• Tip from the excerpt: in radian measure, fractions with denominator 6 have reference angle π/6; denominator 4 → π/4; denominator 3 → π/3.

🔄 Solving basic trigonometric equations

🔄 Finding all angles with a given cosine or sine

Because the Unit Circle repeats every full rotation (2π radians or 360°), solutions come in infinite families.

• Example: cos(θ) = 1/2 means the terminal side intersects the circle at x = 1/2, which happens in Quadrant I (θ = π/3) and Quadrant IV (θ = 5π/3).
• All solutions: θ = π/3 + 2πk or θ = 5π/3 + 2πk for any integer k.
• Example: sin(θ) = −1/2 occurs in Quadrant III (θ = 7π/6) and Quadrant IV (θ = 11π/6), giving θ = 7π/6 + 2πk or θ = 11π/6 + 2πk.
• Example: cos(θ) = 0 (quadrantal angles on the y-axis) gives θ = π/2 + πk for integers k (a shorter way to write both π/2 + 2πk and 3π/2 + 2πk).

🔄 Equivalent solution families

Different-looking formulas can represent the same set of angles.

• Example: θ = 11π/6 + 2πk and θ = −π/6 + 2πk describe the same angles (since 11π/6 and −π/6 are coterminal).
• The excerpt advises: "When in doubt, write it out!"—list a few values to check equivalence.

🌐 Beyond the Unit Circle: circles of radius r

🌐 Extending to any circle centered at the origin

For a circle of radius r (equation x² + y² = r²), if Q(x, y) is the point on the terminal side of θ that lies on this circle, then:

• x = r cos(θ) and y = r sin(θ).
• Conversely, cos(θ) = x/r = x/√(x² + y²) and sin(θ) = y/r = y/√(x² + y²).
• When r = 1 (the Unit Circle), these reduce to the original definitions.

🌐 Example application

• If the terminal side of θ contains the point Q(4, −2), then r = √(16 + 4) = 2√5, so cos(θ) = 4/(2√5) = 2√5/5 and sin(θ) = −2/(2√5) = −√5/5.
• Real-world example: approximating Earth's radius at a given latitude. At 41.628° north, the radius is 3960 cos(41.628°) ≈ 2960 miles.

🔁 Equations of circular motion

🔁 Describing position over time

For an object moving in a circular path of radius r with constant angular velocity ω (starting at (r, 0) when t = 0):

• The angle swept out at time t is θ = ωt.
• The position is (x, y) = (r cos(ωt), r sin(ωt)).
• ω > 0 indicates counter-clockwise motion; ω < 0 indicates clockwise.
• Example: Lakeland Community College rotates with Earth at radius r = 2960 miles and ω = π/12 radians per hour, giving x = 2960 cos(πt/12) and y = 2960 sin(πt/12) (x, y in miles, t in hours).

📐 Right triangle trigonometry

📐 Cosine and sine in terms of triangle sides

For an acute angle θ in a right triangle:

• Let a = length of the side adjacent to θ.
• Let b = length of the side opposite θ.
• Let c = length of the hypotenuse.
• Then cos(θ) = a/c (adjacent over hypotenuse) and sin(θ) = b/c (opposite over hypotenuse).

📐 Solving right triangles

Given one angle and one side, you can find the others.

• Example: A right triangle with a 30° angle and adjacent side 7. The missing angle is 60° (since angles sum to 180°). Using cos(30°) = 7/c gives c = 7/(√3/2) = 14√3/3. Using sin(30°) = b/c gives b = c · (1/2) = 7√3/3.
• Don't confuse: you can use either the Pythagorean Theorem or another trig ratio to find the third side; both methods work.

🔢 Cosine and sine as functions of real numbers

🔢 Extending to all real numbers

By identifying a real number t with the angle θ = t radians, we define cos(t) and sin(t) as functions of real numbers.

• Domain: (−∞, ∞) for both cosine and sine (every real number corresponds to an angle).
• Range: [−1, 1] for both (coordinates on the Unit Circle are between −1 and 1).
• Solving sin(t) = −1/2 is done exactly as solving sin(θ) = −1/2 for θ in radians; the solution is t = 7π/6 + 2πk or t = 11π/6 + 2πk.
• The distinction between "t as a real number" and "θ = t radians" is often blurred in practice; properties developed for angles in radian measure apply equally to real-number inputs.

🧭 Overview

🧠 One-sentence thesis

Sinusoids model a wide range of time-dependent phenomena—from circular motion and daylight variation to harmonic oscillations on springs—by capturing periodic behavior through amplitude, frequency, phase, and vertical shift.

📌 Key points (3–5)

• What a sinusoid models: periodic, oscillating behavior in natural systems (Ferris wheels, springs, daylight hours, moon phases).
• Core parameters: amplitude (maximum displacement from baseline), period (time for one cycle), angular/ordinary frequency (cycles per unit time or per 2π), phase (starting angle), and vertical shift (baseline).
• Harmonic motion: free undamped motion on a spring produces a sinusoid whose angular frequency depends on the spring constant and mass; initial displacement and velocity determine amplitude and phase.
• Common confusion: distinguishing displacement sign conventions—positive vs. negative can mean "below vs. above equilibrium" or "right vs. left of center," depending on context.
• Why it matters: sinusoidal models predict future behavior (e.g., when a Ferris wheel reaches maximum height, when a spring passes through equilibrium) and reveal phenomena like damping, resonance, and beats.

🎢 Modeling circular and periodic motion

🎡 Ferris wheel example

The excerpt models the Giant Wheel at Cedar Point (diameter 128 ft, sits on an 8 ft platform, completes 2 revolutions in 127 seconds).

• Setup: passengers move counter-clockwise on a circle of radius r = 64 ft; the center is 72 ft above ground (8 ft platform + 64 ft radius).
• Baseline formula: the y-coordinate on a circle centered at the origin with angular frequency ω is y = r sin(ωt), where t = 0 corresponds to the point (r, 0).
• Adjustments:
  • Period T = 127/2 seconds (half the total time for one revolution), so angular frequency ω = 2π/T = 4π/127 radians per second.
  • Vertical shift: add 72 to account for the center being 72 ft above ground.
  • Phase shift: passengers start at the lowest point P (not the point Q at angle 0); geometrically this is a shift of −π/2 radians in the angle θ = ωt.
• Final model: h(t) = 64 sin(4π/127 · t − π/2) + 72.
• Interpretation: amplitude 64 matches the radius; phase shift 127/8 ≈ 15.875 seconds is the "time delay" to reach the starting reference point.

Example: At t = 0, passengers are at the lowest point (h = 8 ft); the sinusoid correctly predicts this because sin(−π/2) = −1, giving h = 64(−1) + 72 = 8.

🌞 Daylight hours in Fairbanks

The excerpt fits a sinusoid to monthly daylight data (12 data points over one year).

• Baseline: average of maximum and minimum values: B = (21.8 + 3.3)/2 = 12.55 hours.
• Amplitude: displacement from baseline to maximum: A = 21.8 − 12.55 = 9.25 hours.
• Period: data spans one year (12 months), so T = 12 and ω = 2π/12 = π/6.
• Phase shift: choose the first t where H(t) equals the baseline; here t = 3 (March) gives H ≈ 12.4, close to 12.55, so phase shift = 3 and φ = −3ω = −π/2.
• Manual model: H(t) = 9.25 sin(π/6 · t − π/2) + 12.55.
• Calculator regression: the excerpt notes that a graphing calculator's "SinReg" command may produce a slightly better fit, but no R² value is provided (as with logistic models, this is beyond the course scope).

Don't confuse: the phase shift is not the same as the phase; phase shift = −φ/ω.

🔧 Harmonic motion on springs

🧮 Physics background

Mass: a measure of an object's resistance to straight-line motion (constant everywhere).
Weight: the force gravity exerts on an object (varies with location).

• Conversion: w = mg, where w is weight, m is mass, and g is acceleration due to gravity.
  • English system: g = 32 ft/s², mass in slugs, weight in pounds (lbs).
  • SI system: g = 9.8 m/s², mass in kilograms (kg), weight in Newtons (N).
• Hooke's Law: F = kd, where F is force, k is the spring constant, and d is the stretch distance.

🔩 Equilibrium and displacement

• Equilibrium: the position where the object's weight balances the spring's restorative force.
• Displacement x(t): distance from equilibrium at time t.
  • x(t) = 0: at equilibrium.
  • x(t) < 0: above equilibrium.
  • x(t) > 0: below equilibrium.
• Initial conditions:
  • x₀: initial displacement (same sign convention).
  • v₀: initial velocity; v₀ = 0 means "from rest," v₀ < 0 means upward, v₀ > 0 means downward.

Don't confuse: the sign conventions are inherited from Physics—downward (toward the ground) is positive because it's the "natural" direction without the spring.

📐 Free undamped harmonic motion theorem

Theorem 11.1: If an object of mass m is suspended from a spring with spring constant k, initial displacement x₀, and initial velocity v₀, then the displacement from equilibrium at time t is x(t) = A sin(ωt + φ), where:

• ω = √(k/m)
• A = √(x₀² + (v₀/ω)²)
• A sin(φ) = x₀ and Aω cos(φ) = v₀

• Angular frequency ω: depends only on the spring constant and mass, not on initial conditions.
• Amplitude A: depends on both initial displacement and initial velocity.
• Phase φ: determined by the two equations involving x₀ and v₀.

🧪 Worked example: 64-pound object

The object weighs 64 lbs and stretches the spring 8 ft.

• Spring constant: F = kd → 64 = k · 8 → k = 8 lbs/ft.
• Mass: w = mg → 64 = m · 32 → m = 2 slugs.
• Angular frequency: ω = √(8/2) = 2 rad/s.

Case 1: Released 3 ft below equilibrium from rest (x₀ = 3, v₀ = 0).

• A = √(3² + 0²) = 3.
• A sin(φ) = 3 → 3 sin(φ) = 3 → sin(φ) = 1 → φ = π/2.
• Equation of motion: x(t) = 3 sin(2t + π/2).
• First equilibrium crossing: solve 3 sin(2t + π/2) = 0 → 2t + π/2 = πk → t = π/4 (first positive value) ≈ 0.78 seconds.
• Direction: the graph crosses from positive x (below) to negative x (above), so the object is heading upward—consistent with releasing it below equilibrium.

Case 2: Released 3 ft below equilibrium with upward velocity 8 ft/s (x₀ = 3, v₀ = −8).

• A = √(3² + (−8/2)²) = √(9 + 16) = 5.
• A sin(φ) = 3 → 5 sin(φ) = 3 → sin(φ) = 3/5.
• Aω cos(φ) = −8 → 10 cos(φ) = −8 → cos(φ) = −4/5.
• φ is in Quadrant II: φ = π − arcsin(3/5).
• Equation of motion: x(t) = 5 sin(2t + [π − arcsin(3/5)]).
• Maximum height above equilibrium: amplitude is 5, so the object travels at most 5 ft above equilibrium (x = −5).
• When: solve 5 sin(2t + [π − arcsin(3/5)]) = −5 → t = (1/2) arcsin(3/5) + π/4 ≈ 1.107 seconds.

🌊 Beyond simple harmonic motion

🛑 Damped motion

The excerpt presents x(t) = 10 e^(−t/5) sin(t + π/3).

• Amplitude function: A(t) = 10 e^(−t/5) (not a constant).
• Behavior: as t → ∞, A(t) → 0, so the oscillations shrink toward zero.
• Physical interpretation: friction (e.g., air resistance) gradually slows the motion; the object oscillates forever but with decreasing amplitude.
• Graph: the sinusoid is "enveloped" by the curves y = ±10 e^(−x/5).

📢 Forced motion and resonance

The excerpt presents x(t) = 2(t + 3) sin(2t + π/4).

• Amplitude function: A(t) = 2(t + 3) = 2t + 6 (grows without bound).
• Physical interpretation: the apparatus holding the spring is itself oscillating; when the external frequency matches the spring's natural frequency, resonance occurs and the amplitude grows.
• Graph: the sinusoid is enveloped by y = ±2(x + 3), which diverges as x increases.

🎵 Beats

The excerpt presents x(t) = 5 sin(6t) − 5 sin(8t).

• Period: find the smallest interval where both sin(6t) and sin(8t) complete whole cycles.
  • Ratio of frequencies: 6/8 = 3/4 (reduced).
  • Interpretation: for every 3 cycles of the first, the second makes 4 cycles.
  • Period of x(t): three times the period of 5 sin(6t), which is 2π/6 = π/3, so T = 3 · π/3 = π.
• Sum-to-product identity: x(t) = −10 sin(t) cos(7t).
• Envelope: the lower-frequency factor −10 sin(t) modulates the higher-frequency oscillation cos(7t), creating a "beat" pattern.
• Graph: over [0, 2π], the graph oscillates rapidly but is bounded by y = ±10 sin(x).

Don't confuse: beats arise when two sinusoids have different frequencies; resonance occurs when frequencies match.

📊 Summary table of sinusoid parameters

Parameter	Symbol	Formula / Meaning	Units
Amplitude	|A|	Maximum displacement from baseline	ft or m
Angular frequency	ω	Cycles per 2π units of time	rad/s
Ordinary frequency	f	Cycles per unit time; f = ω/(2π)	Hz (cycles/s)
Period	T	Time for one cycle; T = 1/f = 2π/ω	s
Phase	φ	Angle corresponding to t = 0	radians
Phase shift	−φ/ω	Horizontal "head start"	s
Vertical shift	B	Baseline (center line)	ft or m

Standard form: S(t) = A sin(ωt + φ) + B (for ω > 0).

🧭 Overview

🧠 One-sentence thesis

The Law of Sines provides ratios relating each angle of a triangle to its opposite side, enabling us to solve for unknown sides and angles when we have at least one angle-side opposite pair.

📌 Key points (3–5)

• What the Law of Sines states: For any triangle, the ratio of the sine of each angle to its opposite side is constant across all three angle-side pairs.
• When you can use it: You need at least one complete angle-side opposite pair (an angle and the side directly across from it) to apply the Law of Sines.
• The ambiguous case (ASS): When given two sides and an angle opposite one of them, you may get zero, one, or two valid triangles depending on the side lengths.
• Common confusion: The "Angle-Side-Side" case can produce two different triangles—don't assume there's only one solution without checking the criteria.
• Why it matters: The Law of Sines extends triangle-solving beyond right triangles and provides area formulas using angles and sides.

📐 The Law of Sines formula

📐 Statement of the theorem

The Law of Sines: Given a triangle with angle-side opposite pairs (α, a), (β, b), and (γ, c), the following ratios hold:

sin(α)/a = sin(β)/b = sin(γ)/c

or equivalently: a/sin(α) = b/sin(β) = c/sin(γ)

• An "angle-side opposite pair" means an angle and the side directly across from it in the triangle.
• The excerpt uses Greek letters (α, β, γ) for angles and corresponding lowercase English letters (a, b, c) for their opposite sides.
• This relationship holds for all triangles, not just right triangles.

🔍 How the proof works

The excerpt proves the Law of Sines by breaking it into three cases:

Case	Description	Key technique
All acute angles	Drop an altitude to create two right triangles	Use basic sine ratios from both right triangles
One obtuse angle	Extend an altitude and use supplementary angles	Apply the identity sin(180° - α) = sin(α)
Right triangle	The Law reduces to basic definitions	Falls out from Theorem 10.4 (not detailed here)

• In each case, dropping an altitude creates right triangles where you can apply basic trigonometric ratios.
• The key insight: the same altitude height h can be expressed using sine ratios from different angles, leading to the equal ratios.

🎯 When and how to use the Law of Sines

🎯 The AAS case (Angle-Angle-Side)

What you're given: Two angles and one side (the side is adjacent to only one of the given angles).

• First find the third angle using the fact that angles sum to 180°.
• Then use the Law of Sines with your known angle-side pair to find the other sides.
• This case always produces exactly one triangle.

Example from the excerpt: α = 120°, a = 7 units, β = 45°

• Find γ = 180° - 120° - 45° = 15°
• Use the Law of Sines: b/sin(45°) = 7/sin(120°) to find b
• Use the same approach to find c

🎯 The ASA case (Angle-Side-Angle)

What you're given: Two angles and the side between them.

• Find the third angle (angles sum to 180°).
• Use the Law of Sines with the known side and its opposite angle to find the remaining sides.
• This case also always produces exactly one triangle.

⚠️ The ASS case (Angle-Side-Side) - the ambiguous case

What you're given: One angle, the side opposite it, and one other side.

This is the tricky case—you might get 0, 1, or 2 triangles depending on the measurements.

🔀 The ambiguous case in detail

🔀 Four possible outcomes

The excerpt provides a theorem (Theorem 11.3) that tells you what to expect. Let's say you're given angle α, its opposite side a, and another side c.

First, calculate: h = c sin(α) (this represents a critical height)

Condition	Number of triangles	Why
a < h	Zero triangles	Side a is too short to reach and close the triangle
a = h	Exactly one (right) triangle	Side a just reaches, forming a 90° angle
h < a < c	Two distinct triangles	Side a can "swing" to two different positions
a ≥ c	Exactly one triangle	Side a is long enough that only one configuration works

🔀 Why two triangles can occur

When h < a < c, the Law of Sines gives sin(γ) = c sin(α)/a, which is between 0 and 1.

• There are two angles in the range 0° to 180° with the same sine value: an acute angle γ₀ and its supplement (180° - γ₀).
• Both angles can fit into a triangle with the given angle α.
• This creates two distinct valid triangles.

Don't confuse: If you're given an obtuse angle to start with, you cannot have the two-triangle case (because two obtuse angles cannot fit in one triangle).

🔀 Working through the ambiguous case

Example from the excerpt: α = 30°, a = 3 units, c = 4 units

1. Calculate h = 4 sin(30°) = 2
2. Check: h < a < c? Yes, because 2 < 3 < 4
3. Use Law of Sines: sin(γ)/4 = sin(30°)/3, so sin(γ) = 2/3
4. Find both angles: γ ≈ 41.81° and γ ≈ 138.19°
5. For each γ, find the remaining angle and side to get two complete triangles

🧮 Area formula using the Law of Sines

🧮 The formula

Area of a triangle: Given angle-side opposite pairs (α, a), (β, b), and (γ, c):

A = (1/2)bc sin(α) = (1/2)ac sin(β) = (1/2)ab sin(γ)

• This formula lets you find area when you know two sides and the angle between them.
• Unlike the standard "base times height divided by 2," this works for any triangle without needing to find the height explicitly.
• The excerpt notes: use the formula that incorporates the most given information to minimize propagated error.

🧮 Why this works

The proof uses the same altitude-dropping technique as the Law of Sines proof:

• When you drop an altitude h, you can express h using sine ratios.
• The area is (1/2) × base × height, and the height can be written as h = (side) × sin(angle).
• Substituting this expression gives the formula above.

🗺️ Application: Navigation and bearings

🗺️ Understanding bearing notation

The excerpt introduces a navigation system where directions are given as bearings:

• A bearing describes direction as an acute angle rotated east or west from the north-south line.
• N40°E means: start facing north, rotate 40° toward east (clockwise).
• S50°W means: start facing south, rotate 50° toward west (clockwise).
• Cardinal directions (north, south, east, west) are called "due north," "due south," etc.

Don't confuse: Bearings are not the same as standard position angles (measured counterclockwise from the positive x-axis).

🗺️ Solving bearing problems

Example from the excerpt: Sasquatch Island problem

• Two observation points are 5 miles apart along a straight shoreline.
• From point P, the island is at bearing 30° from shore; from point Q, it's at 45°.
• To solve: Draw a triangle, find all angles using supplementary angles and the 180° sum rule, then apply Law of Sines.

Key technique: Convert bearing information into triangle angles by recognizing supplementary angles and using geometric relationships.

⚠️ Common pitfalls and best practices

⚠️ Minimizing error propagation

The excerpt repeatedly emphasizes: use given data whenever possible, not derived values.

• When you calculate an intermediate value (like an angle), using it in further calculations can compound rounding errors.
• Better approach: Go back to the original given information and use the Law of Sines with those values.

Example: If you're given α and a, and you calculate β, don't use β to find c. Instead, use the original pair (α, a) with the Law of Sines to find c directly.

⚠️ Checking your work in the ambiguous case

When you suspect the ASS case might give two triangles:

1. Calculate h = c sin(α) and compare to the given side lengths.
2. If the Law of Sines gives sin(γ) between 0 and 1, find both possible angles.
3. Check that each angle is compatible with the other given information (e.g., if c > a, then γ > α).
4. Solve completely for both triangles if both are valid.

⚠️ Degree vs. radian mode

The excerpt notes that many real-world applications use degrees, so the examples work in degrees.

• Make sure your calculator is in the correct mode.
• When converting between bearings and standard angles, be careful with the direction of rotation.

🧭 Overview

🧠 One-sentence thesis

The Law of Cosines extends the Pythagorean Theorem to all triangles and enables solving triangles when given two sides and the included angle (SAS) or all three sides (SSS).

📌 Key points (3–5)

• What it solves: Side-Angle-Side (SAS) and Side-Side-Side (SSS) triangle cases, which the Law of Sines cannot handle.
• The formulas: Three equivalent forms relating each side to the other two sides and the opposite angle's cosine.
• Connection to Pythagorean Theorem: When the angle is 90°, the Law of Cosines reduces exactly to the Pythagorean Theorem.
• Common confusion: Unlike sine, cosine distinguishes acute from obtuse angles (positive vs. negative), so the Law of Cosines avoids ambiguity when finding unknown angles.
• Practical outcome: Leads to Heron's Formula for finding triangle area using only the three side lengths.

📐 The Law of Cosines formulas

📐 Three equivalent forms

Law of Cosines: Given a triangle with angle-side opposite pairs (α, a), (β, b), and (γ, c):

• a² = b² + c² − 2bc cos(α)
• b² = a² + c² − 2ac cos(β)
• c² = a² + b² − 2ab cos(γ)

Or solved for the cosine:

• cos(α) = (b² + c² − a²) / (2bc)
• cos(β) = (a² + c² − b²) / (2ac)
• cos(γ) = (a² + b² − c²) / (2ab)

🔍 What each variable means

• Each formula relates one side to the other two sides and the angle opposite that side.
• The angle must be adjacent to (included between) the two known sides in the SAS case.

🧮 How it generalizes Pythagorean Theorem

• When γ = 90°, cos(90°) = 0, so the term −2ab cos(γ) vanishes.
• The formula becomes c² = a² + b², the familiar Pythagorean relation.
• In a broader sense, the Law of Cosines and Pythagorean Theorem are mathematically equivalent—one is the special case of the other.

🛠️ Solving triangles with the Law of Cosines

🛠️ Side-Angle-Side (SAS) case

• Given: two sides and the included angle (e.g., a = 7, c = 2, β = 50°).
• Step 1: Use the Law of Cosines to find the third side (b² = a² + c² − 2ac cos(β)).
• Step 2: Find the remaining angles. Two approaches:
  • Preferred method: Use the Law of Cosines again to find the largest unknown angle first (opposite the longest side). Cosine's sign tells you if the angle is acute (positive) or obtuse (negative).
  • Alternative: Use the Law of Sines, but find the smallest unknown angle first to avoid ambiguity (the smallest angle is guaranteed acute).
• Step 3: Find the last angle by subtraction (sum of angles = 180°) or by another application of the Law of Cosines to minimize rounding error.

🛠️ Side-Side-Side (SSS) case

• Given: all three sides (e.g., a = 4, b = 7, c = 5).
• Step 1: Use the Law of Cosines to find the largest angle first (opposite the longest side). This ensures you catch any obtuse angle.
• Step 2: Use the Law of Cosines (or Law of Sines) to find the remaining angles.
• Advantage: In SSS, you can find all angles using only the original given data, avoiding propagation of rounding errors.

⚠️ Why find the largest angle first

• A triangle can have at most one obtuse angle.
• If there is an obtuse angle, it must be the largest angle (opposite the longest side).
• Cosine is negative for obtuse angles and positive for acute angles, so it unambiguously identifies the angle type.
• Sine is positive for both acute and obtuse angles, so it cannot distinguish between them without additional context.

🌍 Applications

🌍 Real-world distance problems

Example: A researcher measures distances from a point P to two ends of a pond (950 feet and 1000 feet) with a 60° angle between the lines of sight.

• Setup: Two sides and the included angle (SAS).
• Solution: w² = 950² + 1000² − 2(950)(1000) cos(60°) = 952,500, so w ≈ 976 feet.

🌍 Navigation and bearings

• The Law of Cosines handles problems where a path changes direction (e.g., a hiker changes bearing after 1.5 miles).
• Given two legs of a journey and the angle between them, you can find the straight-line distance back to the starting point.

📏 Heron's Formula

📏 What it does

Heron's Formula: For a triangle with sides a, b, c, let s = ½(a + b + c) (the semiperimeter). Then the area A is: A = √[s(s − a)(s − b)(s − c)]

• This formula requires only the three side lengths—no angles or heights needed.

📏 How it's derived

• Start with the area formula A = ½ ab sin(γ) (from the Law of Sines section).
• Square both sides and use the identity sin²(γ) = 1 − cos²(γ).
• Substitute cos(γ) from the Law of Cosines: cos(γ) = (a² + b² − c²) / (2ab).
• Algebraically simplify using difference of squares and perfect square trinomials.
• Recognize that s = (a + b + c)/2 and (s − a) = (b + c − a)/2, etc.
• The final form emerges after factoring and taking the square root.

📏 Example

Given a = 4, b = 7, c = 5:

• s = ½(4 + 7 + 5) = 8
• (s − a) = 4, (s − b) = 1, (s − c) = 3
• A = √(8 · 4 · 1 · 3) = √96 = 4√6 ≈ 9.80 square units.

🔧 Practical tips

🔧 Minimizing rounding error

• When you derive one value (e.g., side b) and then use it to find angles, rounding errors propagate.
• To reduce error:
  • Use the Law of Cosines for all angles (using only the original given data when possible).
  • Carry many decimal places through intermediate steps.
  • Check that approximate angle measures sum to 180° (small discrepancies like 180.01° indicate rounding issues).

🔧 Choosing Law of Cosines vs. Law of Sines

Situation	Preferred method	Reason
SAS or SSS	Law of Cosines	No angle-side opposite pair available; cosine distinguishes acute/obtuse
After finding one side in SAS	Law of Cosines for largest unknown angle	Avoids ambiguity; cosine's sign reveals angle type
After finding one side in SAS	Law of Sines for smallest unknown angle	If using Sines, smallest angle is guaranteed acute

🔧 The ambiguous case revisited

• Applying the Law of Cosines to an Angle-Side-Side (ASS) scenario yields a quadratic equation in the unknown side.
• No positive real solution → no triangle.
• One positive solution → exactly one triangle.
• Two distinct positive solutions → two distinct triangles (the ambiguous case).

🧪 Proof sketch

🧪 Setup

• Place vertex A at the origin, side b along the positive x-axis, so C is at (b, 0).
• Vertex B lies on a circle of radius c at angle α, so B = (c cos(α), c sin(α)).

🧪 Distance formula

• The distance from B to C is side a.
• a = √[(c cos(α) − b)² + (c sin(α))²]
• Square both sides and expand.
• Use cos²(α) + sin²(α) = 1 to simplify.
• Result: a² = b² + c² − 2bc cos(α).

🧪 Why it works for any angle

• The proof holds whether α is acute, right, or obtuse, because the coordinates (c cos(α), c sin(α)) correctly describe any angle in standard position between 0° and 180°.

🧭 Overview

🧠 One-sentence thesis

Polar coordinates provide an alternative to Cartesian coordinates by locating points using a directed distance from a central pole and an angle of rotation from a fixed axis, allowing the same point to be represented by infinitely many coordinate pairs.

📌 Key points (3–5)

• What polar coordinates measure: a directed distance r from the pole (origin) and an angle θ of rotation from the polar axis.
• Multiple representations: unlike Cartesian coordinates, the same point can have infinitely many polar coordinate representations by varying r and θ.
• Negative r values: when r < 0, you move in the opposite direction along the polar axis before rotating.
• Common confusion: the same point can be written as (r, θ) or (−r, θ + π) or (r, θ + 2πk) for any integer k; these all describe the same location in the plane.
• Conversion between systems: polar and Cartesian coordinates are related by x = r cos(θ), y = r sin(θ), and r² = x² + y².

📍 What polar coordinates are

📍 The polar coordinate system

Pole: the origin point in polar coordinates.
Polar axis: a ray from the pole used as the reference direction.
Polar coordinates (r, θ): r represents a directed distance from the pole; θ is a measure of rotation from the polar axis.

• The system measures "how far out" (that's r) and "how far to rotate" (that's θ).
• Contrast with Cartesian: Cartesian coordinates describe rectangular motions (over and up); polar coordinates describe radial and angular motions.

📍 How to plot a point in polar coordinates

Two equivalent approaches:

1. Distance first: start at the pole, move r units along the polar axis, then rotate θ.
2. Angle first: rotate θ from the polar axis, then move outward r units from the pole.

Example: To plot (4, 5π/6), move 4 units along the polar axis, then rotate 5π/6 radians counter-clockwise. Alternatively, rotate 5π/6 first, then move 4 units outward.

🔄 Signed values of r and θ

• When r < 0: move in the opposite direction on the polar axis (away from the usual direction) before rotating.
  • Example: To plot (−3.5, π/4), move 3.5 units in the opposite direction from the polar axis, then rotate π/4. The point ends up on the terminal side of 5π/4, not π/4.
• When θ < 0: rotate clockwise instead of counter-clockwise.
  • Example: (3.5, −3π/4) means rotate 3π/4 clockwise, then move 3.5 units outward.

🔁 Multiple representations of the same point

🔁 Why one point has infinitely many polar coordinates

Unlike Cartesian coordinates where (a, b) and (c, d) are the same point only if a = c and b = d, a single point in polar coordinates can be written in infinitely many ways.

Key insight: You can add full rotations (multiples of 2π) to θ, or flip the sign of r and adjust θ by π.

🔁 Equivalent representations property

Suppose (r, θ) and (r′, θ′) are polar coordinates where r ≠ 0 and r′ ≠ 0 (angles in radians). They determine the same point P if and only if one of the following is true:

• r′ = r and θ′ = θ + 2πk for some integer k
• r′ = −r and θ′ = θ + (2k + 1)π for some integer k

• The first case: same distance, coterminal angles.
• The second case: opposite distance, angles differ by an odd multiple of π.
• Special case: all coordinates of the form (0, θ) represent the pole, regardless of θ.

🔁 Finding alternate representations

To find different representations for a point:

1. For r > 0: choose any angle coterminal with the original θ (add or subtract multiples of 2π).
2. For r < 0: flip the sign of r and adjust θ by adding π (or any odd multiple of π).

Example: The point (2, 240°) can also be written as (2, −120°) (same r, coterminal angle) or (−2, 60°) (opposite r, angle adjusted by 180°).

Don't confuse: (2, 240°) and (−2, 60°) look different as ordered pairs but represent the same location in the plane.

🔄 Converting between polar and Cartesian coordinates

🔄 Conversion formulas

From polar to Cartesian:

• x = r cos(θ)
• y = r sin(θ)

From Cartesian to polar:

• x² + y² = r²
• tan(θ) = y/x (provided x ≠ 0)

These formulas work for all values of r, including negative values.

🔄 Converting points: polar to Cartesian

Straightforward substitution using x = r cos(θ) and y = r sin(θ).

Example: Convert (4, 5π/6) to Cartesian:

• x = 4 cos(5π/6) = 4(−√3/2) = −2√3
• y = 4 sin(5π/6) = 4(1/2) = 2
• Result: (−2√3, 2)

🔄 Converting points: Cartesian to polar

More involved; requires determining both r and θ.

Steps:

1. Plot the point to identify the quadrant.
2. Use r² = x² + y² to find r; choose r ≥ 0 if required.
3. Use tan(θ) = y/x to find the reference angle.
4. Adjust θ based on the quadrant and any restrictions (e.g., 0 ≤ θ < 2π).

Example: Convert (−3, −3) to polar with r ≥ 0 and 0 ≤ θ < 2π:

• r² = (−3)² + (−3)² = 18, so r = 3√2
• tan(θ) = (−3)/(−3) = 1, reference angle π/4
• Point is in Quadrant III, so θ = 5π/4
• Result: (3√2, 5π/4)

Don't confuse: When tan(θ) = y/x, there are infinitely many angles with that tangent value; you must choose the correct one based on the quadrant.

🔄 Special cases

• Points on axes: Use geometric reasoning rather than formulas when x = 0 or y = 0.
  • Example: (0, −3) lies on the negative y-axis, so r = 3 and θ = 3π/2.
• When arctangent is needed: If tan(θ) is not a common value, use θ = arctan(y/x) and adjust for the quadrant.
  • Example: For (−3, 4) in Quadrant II, θ = π − arctan(4/3).

🔀 Converting equations between coordinate systems

🔀 Polar to Cartesian equations

Strategy: Rearrange the polar equation so that expressions like r² = x² + y², r cos(θ) = x, r sin(θ) = y, or tan(θ) = y/x appear, then substitute.

Polar equation	Manipulation	Cartesian result
r = −3	Square both sides: r² = 9	x² + y² = 9
θ = 4π/3	Take tangent: tan(θ) = √3	y = x√3
r = 1 − cos(θ)	Multiply by r: r² = r − r cos(θ); square: r² = (r² + r cos(θ))²	x² + y² = (x² + y² + x)²

Caution: Operations like squaring may introduce extra solutions; verify that the resulting equation describes the same set of points.

🔀 Cartesian to polar equations

Strategy: Replace every x with r cos(θ) and every y with r sin(θ), then simplify using trigonometric identities.

Example: Convert (x − 3)² + y² = 9 to polar:

• Substitute: (r cos(θ) − 3)² + (r sin(θ))² = 9
• Expand: r² cos²(θ) − 6r cos(θ) + 9 + r² sin²(θ) = 9
• Simplify using cos²(θ) + sin²(θ) = 1: r² − 6r cos(θ) = 0
• Factor: r(r − 6 cos(θ)) = 0
• Result: r = 6 cos(θ) (discard r = 0 since it's already included)

Example: Convert y = −x to polar:

• Substitute: r sin(θ) = −r cos(θ)
• Rearrange: r(cos(θ) + sin(θ)) = 0
• Solutions: r = 0 or cos(θ) + sin(θ) = 0
• Geometric insight: y = −x is a line through the origin; θ = −π/4 describes this line (all values of r allowed).
• Result: θ = −π/4

Don't confuse: r = 0 describes only the pole; an equation like θ = −π/4 (with r free) describes an entire line through the pole.

🔀 Why some equations are simpler in one system

• Circles centered at the origin: simple in both systems (x² + y² = a² vs. r = a).
• Lines through the origin: simpler in polar (θ = constant).
• Parabolas like y = x²: simpler in Cartesian; polar form r = sec(θ) tan(θ) is less intuitive.
• Equations like r = 1 − cos(θ): much simpler in polar; Cartesian form x² + y² = (x² + y² + x)² is unwieldy.

The excerpt emphasizes: "relatively nice things in rectangular coordinates, such as y = x², can turn ugly in polar coordinates, and vice-versa."

🧭 Overview

🧠 One-sentence thesis

Graphing polar equations requires tracking how the radius r changes with angle θ, using a θr-plane sketch as a guide to visualize the curve in the xy-plane, and carefully accounting for the fact that any point has infinitely many polar representations when finding intersections.

📌 Key points (3–5)

• The fundamental graphing principle: A point P(r, θ) is on the graph if some representation of P satisfies the equation—not necessarily the first one you write down.
• How to graph systematically: Plot r = f(θ) in the θr-plane first, then translate each interval of θ into the xy-plane by rotating arrows from the origin.
• Negative r values: When r < 0, the curve appears in the opposite direction from where θ points (rotate from the negative x-axis).
• Common confusion—intersection points: Two curves can meet at a point even when no single (r, θ) satisfies both equations simultaneously; you must check multiple representations including (r, θ + 2πk) and (−r, θ + (2k+1)π).
• Why period ≠ graphing interval: The period of f(θ) does not always match the θ-interval needed to trace the complete curve; some curves retrace themselves, others require more than one period.

📐 Basic polar graphs

📐 Constant r: circles centered at the origin

The graph of r = a (where a ≠ 0) is a circle centered at the origin with radius |a|.

• The angle θ is free, so you plot all points at the same distance from the origin.
• Example: r = 4 gives a circle of radius 4; r = −3√2 also gives a circle of radius 3√2 (the sign of r does not change the radius).
• The negative sign means you measure distance in the opposite direction, but you still end up at the same set of points.

📐 Constant θ: lines through the origin

The graph of θ = α is the line containing the terminal side of α when plotted in standard position.

• The radius r is free, so you plot all points along that ray (including negative r, which extends the ray backward through the origin).
• Example: θ = 5π/4 traces out a line through the origin at that angle; θ = −3π/2 traces the y-axis.
• Don't confuse: θ = α is a line, not a ray—negative r values fill in the opposite side.

🎨 Graphing technique: θr-plane to xy-plane

🎨 Why use the θr-plane

• Treat θ as the independent variable and r as the dependent variable, just like y = f(x).
• Plot one cycle of r = f(θ) on the θr-plane (horizontal axis θ, vertical axis r).
• This shows you when r is positive, negative, zero, increasing, or decreasing—critical information for the xy-plane graph.

🎨 Translating to the xy-plane

• Divide the θ-interval into subintervals (e.g., [0, π/2], [π/2, π], etc.).
• For each subinterval, draw arrows in the θr-plane from the θ-axis to the curve.
• In the xy-plane, rotate each arrow through the corresponding angle θ, starting from the origin.
• The length of the arrow is |r|; if r < 0, the arrow points in the opposite direction (rotate from the negative x-axis).

🎨 Example: r = 6 cos(θ)

• As θ runs from 0 to π/2, r decreases from 6 to 0 → the curve starts 6 units out on the positive x-axis and pulls into the origin at the y-axis.
• As θ runs from π/2 to π, r goes from 0 to −6 → the curve is in Quadrant IV (because r < 0), not Quadrant II.
• The complete graph is traced out as θ ranges from 0 to π; beyond that, you retrace the same curve.

🌹 Common polar curves

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

• These produce "petals" radiating from the origin.
• Example: r = 5 sin(2θ) has four petals (two periods of the sine function over [0, 2π]).
• The curve passes through the origin whenever r = 0; these angles are important for sketching.

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

• Shape depends on the ratio of a to b.
• If a = b, you get a cardioid (heart-shaped), e.g., r = 4 − 2 sin(θ).
• If a < b, the limaçon has an inner loop, e.g., r = 2 + 4 cos(θ).
• The inner loop is traced when r < 0.

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

• These are figure-eight curves.
• Because r² appears, you must solve r = ±√(a² cos(2θ)).
• The curve is undefined when the expression under the square root is negative.
• Example: r² = 16 cos(2θ) is defined only when cos(2θ) ≥ 0, i.e., on [0, π/4] and [3π/4, π] (and their extensions).

🔍 Finding intersection points

🔍 Why intersections are tricky

• A point P can have infinitely many representations: (r, θ), (r, θ + 2π), (−r, θ + π), etc.
• Two curves can meet at P even if no single (r, θ) satisfies both equations.
• You must check multiple representations systematically.

🔍 Step 1: Sketch and check the origin

• Graph both curves to see how many intersections to expect.
• The origin (pole) can be represented as (0, θ) for any θ.
• Check if each curve passes through the origin (set r = 0 and solve for θ).
• If both curves pass through the origin but at different angles θ, the origin is still an intersection point.

🔍 Step 2: Equate the equations

• Assume a point P has a representation (r, θ) that satisfies both equations.
• Set the two expressions for r equal and solve for θ.
• Substitute back to find the corresponding r values.
• Example: For r = 2 sin(θ) and r = 2 − 2 sin(θ), equate to get 2 sin(θ) = 2 − 2 sin(θ), so sin(θ) = 1/2, giving θ = π/6 or 5π/6.

🔍 Step 3: Check (r, θ + 2πk)

• A point might have one representation (r, θ) on the first curve and another (r, θ + 2πk) on the second.
• Substitute θ + 2πk into one of the equations (not both) and see if you get new solutions.
• Often this reduces to the same equation (because trig functions are periodic), yielding no new points.

🔍 Step 4: Check (−r, θ + (2k+1)π)

• A point might have representation (r, θ) on one curve and (−r, θ + (2k+1)π) on the other.
• Substitute −r for r and θ + (2k+1)π for θ into one equation.
• Simplify using trig identities (e.g., cos(θ + π) = −cos(θ)).
• Example: For r = 3 and r = 6 cos(2θ), substituting gives −r = 6 cos(2(θ + π)) = 6 cos(2θ), so r = −6 cos(2θ). Equate with r = 3 to get cos(2θ) = −1/2, yielding four more intersection points.

🔍 Don't confuse: same point vs. same representation

• (3, π/3) and (−3, π/3) are different points.
• (3, π/3) and (3, π/3 + 2π) are the same point.
• (3, π/3) and (−3, 4π/3) are the same point (because 4π/3 = π/3 + π).

🗺️ Describing regions

🗺️ Set-builder notation for polar regions

• A region is described by inequalities on r and θ.
• Example: {(r, θ) | 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π/2} is the set of all points whose distance from the origin is between 0 and 5 sin(2θ) as θ ranges from 0 to π/2.
• This describes one petal of the rose r = 5 sin(2θ).

🗺️ Interpreting inequalities on r

• 0 ≤ r ≤ f(θ) means "from the origin out to the curve r = f(θ)."
• a ≤ r ≤ b means "between the circle r = a and the circle r = b."
• f(θ) ≤ r ≤ 0 means "from the curve r = f(θ) (where r < 0) to the origin"—this traces the inner loop of a limaçon.

🗺️ Union of regions

• Use ∪ to combine two regions.
• Example: {(r, θ) | 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π/6} ∪ {(r, θ) | 0 ≤ r ≤ 2 − 2 sin(θ), π/6 ≤ θ ≤ π/2} describes a region that switches from one curve to another at θ = π/6 (the intersection angle).

🗺️ Example: inner loop of a limaçon

• For r = 2 + 4 cos(θ), the inner loop occurs when r ≤ 0.
• Solve 2 + 4 cos(θ) = 0 to find cos(θ) = −1/2, so θ = 2π/3 or 4π/3.
• The inner loop is {(r, θ) | 2 + 4 cos(θ) ≤ r ≤ 0, 2π/3 ≤ θ ≤ 4π/3}.

⚠️ Important reminders

⚠️ Period vs. graphing interval

• The period of f(θ) tells you when the function values repeat.
• The graphing interval tells you when the curve in the xy-plane repeats.
• These are not always the same!
• Example: r = 6 cos(θ) has period 2π, but the complete graph is traced as θ ranges from 0 to π (half the period).
• Example: r = 5 sin(2θ) has period π, but the complete graph requires θ ∈ [0, 2π] (two periods).

⚠️ When r crosses zero

• The curve passes through the origin when r = 0.
• The line θ = α (where r = 0) acts as a "guide" for the curve as it enters or exits the origin.
• A theorem from Calculus says the curve "hugs" this line near the origin.

⚠️ Symmetry

• Many polar curves exhibit symmetry about the x-axis, y-axis, or origin.
• Symmetry can simplify graphing, but testing for symmetry in polar coordinates is more subtle than in rectangular coordinates.
• The excerpt mentions that even if f(θ) is not even, the graph of r = f(θ) can still be symmetric about the x-axis (e.g., r = 3 sin(θ/2)).

🧭 Overview

🧠 One-sentence thesis

Conic sections can be unified through rotation of axes (which eliminates troublesome xy-terms) and through polar equations (which express all conics via eccentricity and a focus-directrix relationship).

📌 Key points (3–5)

• Rotation of axes: substituting rotated coordinates x′, y′ into equations with xy-terms can eliminate the cross-product term and reveal the conic's standard form.
• Discriminant test: the quantity B² − 4AC determines the conic type (hyperbola if > 0, parabola if = 0, ellipse/circle if < 0) without rotating axes.
• Polar form unifies conics: all conics share the form r = ed/(1 − e cos(θ − φ)), where e is eccentricity (e < 1 ellipse, e = 1 parabola, e > 1 hyperbola).
• Common confusion: the coefficients A and C in Ax² + Bxy + Cy² do not directly tell you the conic type when B ≠ 0; you must check the discriminant or rotate axes first.
• Focus-directrix definition: every conic is the locus of points P such that (distance to focus F)/(distance to directrix L) = e, reconciling the "new" and "old" definitions from Chapter 7.

---

🔄 Rotation of axes

🔄 Why rotate axes

• An equation like y = 2/x or 21x² + 10xy√3 + 31y² = 144 contains an xy cross-product term that makes graphing difficult.
• Rotating the x- and y-axes counter-clockwise through an angle θ produces new axes x′ and y′; the same point P has coordinates (x, y) in the old system and (x′, y′) in the new system.
• The goal: choose θ so that the equation in x′, y′ has no x′y′ term, revealing a standard conic form (e.g., ellipse, hyperbola, parabola).

📐 Rotation formulas (Theorem 11.9)

Rotation of Axes: If the positive x- and y-axes are rotated counter-clockwise through angle θ to produce x′- and y′-axes, then the coordinates P(x, y) and P(x′, y′) are related by:

• x = x′ cos(θ) − y′ sin(θ)
• y = x′ sin(θ) + y′ cos(θ)
  and conversely:
• x′ = x cos(θ) + y sin(θ)
• y′ = −x sin(θ) + y cos(θ)

• These formulas come from converting both coordinate systems to polar coordinates with the same r and using sum formulas for sine and cosine.
• The excerpt derives the inverse formulas using matrix algebra: the rotation matrix A(θ) has determinant 1, so it is invertible.
• Example: rotating through θ = π/3, the point (2, −4) becomes approximately (−2.46, −3.73) in the new system; substituting back confirms the original coordinates.

🎯 Choosing the rotation angle (Theorem 11.10)

• Given Ax² + Bxy + Cy² + Dx + Ey + F = 0 with B ≠ 0, you can eliminate the x′y′ term by choosing θ so that cot(2θ) = (A − C)/B.
• This formula arises by substituting the rotation equations into the original equation, collecting the coefficient of x′y′, and setting it to zero.
• Example: for 21x² + 10xy√3 + 31y² = 144, we have A = 21, B = 10√3, C = 31, so cot(2θ) = (21 − 31)/(10√3) = −1/√3, giving 2θ = 2π/3, hence θ = π/3.
• After rotation, the equation simplifies to (x′)²/4 + (y′)²/9 = 1, an ellipse.

🧮 Working through the substitution

• Substitute x = x′ cos(θ) − y′ sin(θ) and y = x′ sin(θ) + y′ cos(θ) into the original equation.
• Expand and collect like terms: the coefficient of (x′)², the coefficient of (y′)², and (crucially) the coefficient of x′y′.
• The excerpt shows that x² = (x′)² cos²(θ) − 2x′y′ cos(θ)sin(θ) + (y′)² sin²(θ), and similarly for xy and y².
• Combining these with A, B, C yields the x′y′ coefficient: −2A cos(θ)sin(θ) + B(cos²(θ) − sin²(θ)) + 2C cos(θ)sin(θ).
• Using double-angle identities (sin(2θ) = 2 sin(θ)cos(θ), cos(2θ) = cos²(θ) − sin²(θ)), this simplifies to B cos(2θ) − (A − C)sin(2θ).
• Setting this to zero gives cot(2θ) = (A − C)/B.

---

🔍 Discriminant and conic classification

🔍 The discriminant B² − 4AC (Theorem 11.11)

Discriminant test: For a non-degenerate conic Ax² + Bxy + Cy² + Dx + Ey + F = 0:

• If B² − 4AC > 0, the graph is a hyperbola.
• If B² − 4AC = 0, the graph is a parabola.
• If B² − 4AC < 0, the graph is an ellipse or circle.

• This quantity B² − 4AC is called the discriminant of the conic section.
• It allows you to classify the conic without rotating axes or completing the square.
• Example: for 21x² + 10xy√3 + 31y² = 144, B² − 4AC = (10√3)² − 4(21)(31) = 300 − 2604 = −2304 < 0, so it is an ellipse.

🧩 Why the discriminant works

• The excerpt proves that after rotating axes to eliminate the x′y′ term, the new equation has coefficients A′ and C′ such that B² − 4AC = −4A′C′.
• In the rotated (standard) form, the product A′C′ determines the conic type (from Exercise 34 in Section 7.5): A′C′ > 0 → ellipse, A′C′ = 0 → parabola, A′C′ < 0 → hyperbola.
• Since B² − 4AC has the opposite sign of A′C′, the discriminant test follows.
• The proof involves heavy algebra: substituting rotation formulas, using power-reduction and double-angle identities, and the condition cot(2θ) = (A − C)/B to simplify the product 4A′C′ = 4AC − B².

⚠️ Don't confuse: coefficients vs discriminant

• In Chapter 7 (without xy-terms), you could tell the conic type from A and C alone: if both positive and equal → circle, both positive and unequal → ellipse, opposite signs → hyperbola, one zero → parabola.
• When B ≠ 0, those rules break down: Example 11.6.2 shows 5x² + 26xy + 5y² (both A and C positive and equal) is actually a hyperbola, and 16x² + 24xy + 9y² (both positive) is a parabola.
• Always use the discriminant B² − 4AC when an xy-term is present.

---

🌀 Polar form of conics

🌀 Unified definition via focus and directrix (Definition 11.1)

Conic section: Given a fixed line L (directrix), a point F not on L (focus), and a positive number e (eccentricity), a conic section is the set of all points P such that
(distance from P to F) / (distance from P to L) = e.

• This definition unifies parabolas, ellipses, and hyperbolas under one framework.
• For a parabola, e = 1 (equidistant from focus and directrix).
• For an ellipse, 0 < e < 1 (closer to focus than to directrix on average).
• For a hyperbola, e > 1 (farther from focus than from directrix on average).

📏 Deriving the polar equation

• Place the focus F at the origin and the directrix at x = −d (a vertical line to the left).
• For a point P(r, θ) with r > 0, the distance to F is r, and the distance to the line x = −d is d + r cos(θ).
• Setting e = r / (d + r cos(θ)) and solving for r gives r = ed / (1 − e cos(θ)).
• Converting this back to rectangular coordinates and comparing with Chapter 7 formulas confirms that the "new" focus, directrix, and eccentricity match the "old" definitions.

🎨 Four standard orientations (Theorem 11.12)

• Depending on whether the directrix is x = −d, x = d, y = −d, or y = d, the polar equation takes one of four forms:
  • r = ed / (1 − e cos(θ)) (directrix x = −d)
  • r = ed / (1 + e cos(θ)) (directrix x = d)
  • r = ed / (1 − e sin(θ)) (directrix y = −d)
  • r = ed / (1 + e sin(θ)) (directrix y = d)
• In all cases, (0, 0) is a focus and e is the eccentricity.
• The directrix is always perpendicular to the major axis (ellipse) or transverse axis (hyperbola); for parabolas, knowing the focus and directrix tells you which way it opens.

📊 Lengths and dimensions (Theorem 11.12)

Conic type	Condition	Major/transverse axis length	Minor/conjugate axis length	Focal diameter (parabola)
Ellipse	0 < e < 1	2ed / (1 − e²)	2ed / √(1 − e²)	—
Parabola	e = 1	—	—	2d
Hyperbola	e > 1	2ed / (e² − 1)	2ed / √(e² − 1)	—

• Example: r = 12 / (3 − cos(θ)) rewrites as r = 4 / (1 − (1/3)cos(θ)), so e = 1/3, ed = 4, d = 12, directrix x = −12.
• Major axis length = 2(4) / (1 − 1/9) = 9; minor axis length = 2(4) / √(1 − 1/9) = 6√3.
• Vertices at r(0) = 6 and r(π) = 3, corresponding to rectangular points (6, 0) and (−3, 0); center at (3/2, 0).

🔄 Rotated polar conics (Theorem 11.13)

• Replacing θ with (θ − φ) rotates the graph counter-clockwise by angle φ.
• General form: r = ℓ / (1 − e cos(θ − φ)), where ℓ > 0, e ≥ 0, and φ is the rotation angle.
• If e = 0, the graph is a circle centered at (0, 0) with radius ℓ.
• If e ≠ 0, the focus is at (0, 0) and the directrix contains the point with polar coordinates (−d, φ), where d = ℓ/e.
• All length formulas from Theorem 11.12 remain valid because rotations preserve distances.
• Example: r = 4 / (1 − sin(θ − π/4)) is the parabola r = 4 / (1 − sin(θ)) rotated counter-clockwise by π/4.

---

🧪 Worked examples

🧪 Example: rotating to eliminate xy

• Problem: rotate axes to graph 5x² + 26xy + 5y² − 16x√2 + 16y√2 − 104 = 0.
• Solution: A = 5, B = 26, C = 5, so cot(2θ) = (5 − 5)/26 = 0, giving θ = π/4.
• Rotation equations: x = (x′ − y′)/√2, y = (x′ + y′)/√2.
• After substitution and simplification, the equation becomes 18(x′)² − 8(y′)² + 32y′ − 104 = 0, or (x′)²/4 − (y′ − 2)²/9 = 1.
• This is a hyperbola centered at (0, 2) in x′y′-coordinates, opening along the x′-axis.

🧪 Example: using the discriminant

• For 21x² + 10xy√3 + 31y² = 144: B² − 4AC = (10√3)² − 4(21)(31) = −2304 < 0 → ellipse.
• For 5x² + 26xy + 5y² − 16x√2 + 16y√2 − 104 = 0: B² − 4AC = 26² − 4(5)(5) = 576 > 0 → hyperbola.
• For 16x² + 24xy + 9y² + 15x − 20y = 0: B² − 4AC = 24² − 4(16)(9) = 0 → parabola.

🧪 Example: graphing polar conics

• r = 4 / (1 − sin(θ)): e = 1 (parabola), ed = 4, d = 4, directrix y = −4.
• Vertex at (0, −2) in rectangular coordinates, opens upward, focal diameter 2d = 8.
• r = 12 / (3 − cos(θ)) = 4 / (1 − (1/3)cos(θ)): e = 1/3 (ellipse), ed = 4, d = 12, directrix x = −12.
• Vertices at (6, 0) and (−3, 0), center (3/2, 0), major axis length 9, minor axis length 6√3.
• r = 6 / (1 + 2sin(θ)): e = 2 (hyperbola), ed = 6, d = 3, directrix y = 3.
• Vertices at (0, 2) and (0, 6), center (0, 4), transverse axis length 4, conjugate axis length 4√3, asymptotes y = ±(√3/3)x + 4.

---

🔗 Connections and notes

🔗 Reconciling old and new definitions

• The "new" focus-directrix-eccentricity definition (Definition 11.1) matches the "old" definitions from Chapter 7 for parabolas (Definition 7.3), ellipses (Definition 7.5), and hyperbolas (Exercise 31 in Section 7.5).
• The excerpt verifies this by converting r = ed / (1 − e cos(θ)) to rectangular form and checking that the resulting ellipse/hyperbola/parabola has the correct focus, directrix, and eccentricity according to Chapter 7 formulas.

🔗 Rotation matrix

• The matrix A(θ) = [cos(θ), −sin(θ); sin(θ), cos(θ)] is called a rotation matrix.
• It has determinant 1, so it is invertible: A(θ)⁻¹ = A(−θ) (using even/odd identities for cosine and sine).
• Geometrically, A(−θ) rotates points clockwise by θ, undoing the counter-clockwise rotation by θ.

🔗 Why sin(2θ) ≠ 0 is safe

• When deriving cot(2θ) = (A − C)/B, the excerpt divides by sin(2θ).
• If sin(2θ) = 0, then B cos(2θ) = 0; since B ≠ 0, this forces cos(2θ) = 0.
• No angle can have both sin(2θ) = 0 and cos(2θ) = 0 simultaneously, so sin(2θ) ≠ 0 is guaranteed.
• Geometrically, sin(2θ) = 0 would mean the axes are not actually rotated (or rotated by π), which is a degenerate case.

🔗 Extending eccentricity to circles

• Theorem 11.13 includes e = 0, which gives r = ℓ (a circle centered at the origin with radius ℓ).
• This extends the concept of eccentricity to circles, completing the hierarchy: circle (e = 0), ellipse (0 < e < 1), parabola (e = 1), hyperbola (e > 1).

🧭 Overview

🧠 One-sentence thesis

The polar form of a complex number expresses it in terms of its distance from the origin (modulus) and angle from the positive real axis (argument), which dramatically simplifies multiplication, division, and especially computing powers and roots of complex numbers.

📌 Key points (3–5)

• What polar form is: Every complex number z = a + bi can be written as |z| cis(θ) = |z|[cos(θ) + i sin(θ)], where |z| is the modulus (distance from origin) and θ is an argument (angle).
• Why polar form matters: Multiplying complex numbers in polar form means multiplying moduli and adding arguments; dividing means dividing moduli and subtracting arguments; raising to powers means raising the modulus to that power and multiplying the argument by that power (DeMoivre's Theorem).
• Finding roots becomes systematic: To find all n distinct nth roots of a complex number, take the nth root of the modulus and divide the argument by n, then rotate by 2π/n radians successively to get all n roots equally spaced around the complex plane.
• Common confusion – arguments vs principal argument: A complex number has infinitely many arguments (all coterminal angles), but only one principal argument Arg(z) in the interval (−π, π]; arg(z) is the set of all arguments, while Arg(z) is the single value in that restricted interval.
• Geometric interpretation: Multiplying by a complex number w scales distances by |w| and rotates by Arg(w); division scales by 1/|w| and rotates by −Arg(w).

📐 The complex plane and rectangular form

📐 Complex plane setup

• The complex plane replaces the usual x-y coordinate system with a real axis (horizontal) and an imaginary axis (vertical, marked in increments of i).
• Every complex number z = a + bi corresponds to the point (a, b) in this plane.
• The expression z = a + bi is called the rectangular form of z.

Real part: Re(z) = a, the horizontal coordinate.

Imaginary part: Im(z) = b, the vertical coordinate (note: b is a real number; the "imaginary part" does not include the i).

📍 Why this matters

• Visualizing complex numbers as points or vectors in the plane makes geometric operations (rotation, scaling) intuitive.
• The rectangular form is natural for addition and subtraction but cumbersome for multiplication, division, and powers.

🧮 Modulus and argument

🧮 Modulus definition and properties

Modulus: |z| is the distance from z to 0 in the complex plane; if (r, θ) is a polar representation of (a, b) with r ≥ 0, then |z| = r.

• From the distance formula: |z| = √(a² + b²) = √(Re(z)² + Im(z)²).
• The modulus is always non-negative: |z| ≥ 0, and |z| = 0 if and only if z = 0.
• Product rule: |zw| = |z||w|; the modulus of a product is the product of the moduli.
• Power rule: |z^n| = |z|^n for any natural number n.
• Quotient rule: |z/w| = |z|/|w| (provided w ≠ 0).
• The notation |z| is justified because for real numbers, modulus coincides with absolute value.

Example: For z = √3 − i, we have |z| = √((√3)² + (−1)²) = √4 = 2.

🔄 Argument definition and properties

Argument: If (r, θ) is a polar representation of the point (a, b) corresponding to z, then θ is an argument of z; the set of all arguments is denoted arg(z).

Principal argument: If z ≠ 0 and −π < θ ≤ π, then θ is the principal argument, written Arg(z).

• For z ≠ 0, all arguments are coterminal (differ by integer multiples of 2π): arg(z) = {Arg(z) + 2πk | k is an integer}.
• Only one argument lies in (−π, π], and that is the principal argument.
• If Re(z) ≠ 0: tan(θ) = Im(z)/Re(z) for any θ ∈ arg(z); use the quadrant of z to determine the correct angle.
• If Re(z) = 0 and Im(z) > 0: arg(z) = {π/2 + 2πk | k is an integer}.
• If Re(z) = 0 and Im(z) < 0: arg(z) = {−π/2 + 2πk | k is an integer}.
• If z = 0: arg(0) = (−∞, ∞) (all angles work), and Arg(0) is undefined.

Example: For z = √3 − i (Quadrant IV), tan(θ) = −1/√3, so θ = −π/6 + 2πk; the principal argument is Arg(z) = −π/6.

⚠️ Don't confuse: argument vs principal argument

• arg(z) is a set of infinitely many angles; Arg(z) is a single number (or undefined for z = 0).
• Write "θ ∈ arg(z)" (θ is in the set of arguments), not "θ = arg(z)."

🎯 Polar form definition and conversion

🎯 Defining polar form

• From polar coordinates (r, θ) with r ≥ 0 for the point (a, b), we have a = r cos(θ) and b = r sin(θ).
• Substituting into z = a + bi gives z = r cos(θ) + r sin(θ) i = r[cos(θ) + i sin(θ)].
• The expression cos(θ) + i sin(θ) is abbreviated cis(θ).

Polar form: If z is a complex number and θ ∈ arg(z), then z = |z| cis(θ) = |z|[cos(θ) + i sin(θ)] is a polar form for z.

• Since there are infinitely many choices for θ, there are infinitely many polar forms for z (all equivalent).

🔄 Converting rectangular to polar

1. Compute |z| = √(a² + b²).
2. Find an argument θ using tan(θ) = b/a (if a ≠ 0) and the quadrant of z.
3. Write z = |z| cis(θ).

Example: For z = −2 + 4i, |z| = √(4 + 16) = 2√5; since z is in Quadrant II and tan(θ) = −2, we have θ = π − arctan(2), so z = 2√5 cis(π − arctan(2)).

🔄 Converting polar to rectangular

1. Expand cis(θ) = cos(θ) + i sin(θ).
2. Multiply by |z| and simplify using known values or identities.

Example: z = 4 cis(2π/3) = 4[cos(2π/3) + i sin(2π/3)] = 4[−1/2 + i(√3/2)] = −2 + 2i√3.

⚙️ Operations in polar form

⚙️ Product rule

If z = |z| cis(α) and w = |w| cis(β), then:

zw = |z||w| cis(α + β)

• Multiply the moduli and add the arguments.
• Proof uses the sum identities for cosine and sine after expanding cis(α) cis(β).

Example: z = 4 cis(π/6) and w = 2 cis(2π/3) gives zw = 8 cis(π/6 + 2π/3) = 8 cis(5π/6).

⚙️ Quotient rule

If z = |z| cis(α) and w = |w| cis(β) with w ≠ 0, then:

z/w = (|z|/|w|) cis(α − β)

• Divide the moduli and subtract the arguments.
• Proof uses the complex conjugate of the denominator and difference identities.

Example: z = 4 cis(π/6) and w = 2 cis(2π/3) gives z/w = 2 cis(π/6 − 2π/3) = 2 cis(−π/2).

🔥 Power rule (DeMoivre's Theorem)

For any natural number n:

z^n = |z|^n cis(nθ)

• Raise the modulus to the nth power and multiply the argument by n.
• Proof by induction using the product rule.

Example: w = 2 cis(2π/3), so w⁵ = 2⁵ cis(5 · 2π/3) = 32 cis(10π/3) = 32 cis(4π/3) (using coterminal angles).

💡 Why polar form is powerful

• Computing (−1 + i√3)⁵ by expanding with the Binomial Theorem is tedious; in polar form it's one step.
• Division like (2√3 + 2i)/(−1 + i√3) requires rationalizing the denominator in rectangular form but is straightforward in polar form.

🌀 Geometric interpretation of operations

🌀 Multiplication as scaling and rotation

• Multiplying z by w scales the distance from 0 to z by the factor |w|.
• Then rotates the result counter-clockwise by Arg(w) radians (if Arg(w) > 0).

Example: If z = 4 cis(π/6) and w = 2 cis(2π/3), first double the distance from 0 to z (getting 8 cis(π/6)), then rotate counter-clockwise by 2π/3 to arrive at zw = 8 cis(5π/6).

🌀 Division as shrinking and rotation

• Dividing z by w shrinks the distance from 0 to z by the factor |w| (multiplies by 1/|w|).
• Then rotates clockwise by Arg(w) radians (equivalently, rotates counter-clockwise by −Arg(w)).

Example: z/w with z = 4 cis(π/6) and w = 2 cis(2π/3) means halve the distance (getting 2 cis(π/6)), then rotate clockwise by 2π/3 to get 2 cis(−π/2).

⚠️ Don't confuse: scaling vs translating

• Multiplication scales and rotates around the origin; it does not translate (shift) the point.
• The origin (z = 0) is fixed under multiplication by any non-zero complex number.

🌱 Finding nth roots

🌱 Definition of nth roots

nth root: w is an nth root of z if w^n = z for some natural number n.

• Unlike real numbers, complex numbers have n distinct nth roots (for z ≠ 0).
• Example: The cube roots of 8 are 2, −1 + i√3, and −1 − i√3 (found by solving w³ = 8).

🌱 Formula for nth roots

Let z ≠ 0 with polar form z = r cis(θ). The n distinct nth roots of z are:

w_k = ⁿ√r cis(θ/n + (2π/n)k) for k = 0, 1, 2, ..., n−1.

• Take the nth root of the modulus: ⁿ√r.
• Divide the argument by n: θ/n.
• Add 2π/n successively to get all n roots.

Example: To find the cube roots of z = 8 = 8 cis(0), we have ³√8 = 2 and θ/3 = 0, so:

• w₀ = 2 cis(0) = 2
• w₁ = 2 cis(2π/3) = −1 + i√3
• w₂ = 2 cis(4π/3) = −1 − i√3

🌀 Geometric pattern of roots

• All n roots have the same modulus: ⁿ√r.
• They are equally spaced around a circle of radius ⁿ√r, separated by angles of 2π/n radians.
• This creates a regular n-gon centered at the origin.

Example: The four fourth roots of −16 are equally spaced π/2 radians apart around a circle of radius 2.

⚠️ Don't confuse: principal root vs all roots

• For real numbers, ⁿ√x denotes the principal nth root (one specific value).
• For complex numbers, we find all n distinct nth roots; there is no single "principal" nth root in general.

🔢 Special case: roots of unity

🔢 What are roots of unity

nth roots of unity: The n complex nth roots of z = 1.

• Since 1 = 1 cis(0), the nth roots of unity are w_k = cis(2πk/n) for k = 0, 1, ..., n−1.
• All have modulus 1 and are equally spaced around the unit circle.

Example: The five fifth roots of 1 are cis(0) = 1, cis(2π/5), cis(4π/5), cis(6π/5), and cis(8π/5).

🔢 Properties of roots of unity

• w = 1 is always an nth root of unity.
• If w_j and w_k are nth roots of unity, so is their product w_j w_k.
• For every nth root of unity w_j, there exists another nth root w_j' such that w_j w_j' = 1 (they are multiplicative inverses).

💡 Why roots of unity matter

• They form the vertices of a regular n-gon inscribed in the unit circle.
• They have algebraic structure (they form a group under multiplication).
• Euler's formula connects them to the exponential function: e^(iθ) = cis(θ), so the nth roots of unity are e^(2πik/n).

---

Note: The excerpt includes worked examples showing conversions between rectangular and polar forms, computing products/quotients/powers in polar form, and finding specific nth roots. All formulas are proven using trigonometric identities (sum/difference formulas, Pythagorean identity) and mathematical induction (for the power rule). The geometric visualizations emphasize that polar form reveals the "scaling and rotation" nature of complex multiplication.

🧭 Overview

🧠 One-sentence thesis

Vectors combine magnitude (size) and direction into a single mathematical object that can be manipulated algebraically through component form, enabling the modeling of real-world phenomena like velocity, force, and displacement.

📌 Key points (3–5)

• What vectors represent: quantities that have both magnitude (length) and direction, unlike scalars which have only magnitude.
• Component form vs geometric form: vectors can be written as ordered pairs ⟨v₁, v₂⟩ (component form) or as directed line segments with initial and terminal points.
• Vector operations work component-wise: addition, subtraction, and scalar multiplication all operate on corresponding components separately.
• Common confusion—magnitude vs direction: the magnitude ‖v‖ is a scalar (non-negative number), while the direction v̂ is itself a unit vector; together they reconstruct the original vector as v = ‖v‖v̂.
• Why vectors matter: they model forces, velocities, and other directed quantities in physics and engineering, allowing algebraic solutions to geometric problems.

📐 Defining vectors and component form

📐 What is a vector

A vector is a mathematical object characterized by two properties: magnitude (length) and direction.

• Geometrically: a directed line segment with an arrow indicating direction.
• The initial point (tail) is where the vector starts; the terminal point (head) is where it ends.
• Notation: v⃗ denotes "the vector v"; if drawn from point P to point Q, we write v⃗ = PQ⃗.
• Key idea: any directed segment with the same length and direction represents the same vector, regardless of where it starts.

📦 Component form

The component form of a vector v⃗ with initial point P(x₀, y₀) and terminal point Q(x₁, y₁) is v⃗ = ⟨x₁ − x₀, y₁ − y₀⟩.

• The first number (x₁ − x₀) is the x-component; the second (y₁ − y₀) is the y-component.
• Component form captures "how far right/left" and "how far up/down" the vector moves.
• Example: if P = (1, 2) and Q = (4, 6), then v⃗ = ⟨4 − 1, 6 − 2⟩ = ⟨3, 4⟩.
• Two vectors are equal if and only if their corresponding components are equal: ⟨v₁, v₂⟩ = ⟨v'₁, v'₂⟩ means v₁ = v'₁ and v₂ = v'₂.

🔄 Standard position

• A vector is in standard position when its initial point is at the origin (0, 0).
• If v⃗ = ⟨v₁, v₂⟩ is in standard position, its terminal point is exactly (v₁, v₂).
• Standard position simplifies converting between rectangular and polar representations.

➕ Vector addition and subtraction

➕ Geometric addition (resultant vector)

• To add v⃗ + w⃗ geometrically: plot v⃗, then plot w⃗ starting from v⃗'s terminal point; the sum is the vector from v⃗'s initial point to w⃗'s terminal point.
• The sum v⃗ + w⃗ is called the resultant vector—the "net result" of moving along v⃗ then w⃗.
• Visualize: v⃗ and w⃗ form two sides of a parallelogram; v⃗ + w⃗ is one diagonal.

➕ Component-wise addition

For v⃗ = ⟨v₁, v₂⟩ and w⃗ = ⟨w₁, w₂⟩, the sum is v⃗ + w⃗ = ⟨v₁ + w₁, v₂ + w₂⟩.

• Add corresponding components separately.
• Example: ⟨3, 4⟩ + ⟨1, −2⟩ = ⟨3 + 1, 4 + (−2)⟩ = ⟨4, 2⟩.

➖ Subtraction and additive inverse

• The additive inverse of v⃗ = ⟨v₁, v₂⟩ is −v⃗ = ⟨−v₁, −v₂⟩.
• Geometrically: −v⃗ has the same length as v⃗ but points in the opposite direction.
• Subtraction: v⃗ − w⃗ = v⃗ + (−w⃗) = ⟨v₁ − w₁, v₂ − w₂⟩.
• Geometric interpretation: v⃗ − w⃗ is the vector from w⃗'s terminal point to v⃗'s terminal point (when both are drawn from the same initial point).

🔁 Properties of vector addition

Property	Statement
Commutative	v⃗ + w⃗ = w⃗ + v⃗
Associative	(u⃗ + v⃗) + w⃗ = u⃗ + (v⃗ + w⃗)
Identity	v⃗ + 0⃗ = v⃗ (where 0⃗ = ⟨0, 0⟩)
Inverse	v⃗ + (−v⃗) = 0⃗

• The zero vector 0⃗ = ⟨0, 0⟩ acts as a point; its direction is undefined but it serves as the additive identity.
• These properties mirror real-number arithmetic.

✖️ Scalar multiplication

✖️ Definition and geometric meaning

For a scalar k (a real number) and vector v⃗ = ⟨v₁, v₂⟩, scalar multiplication is kv⃗ = ⟨kv₁, kv₂⟩.

• Multiply each component by the scalar.
• Geometric effect:
  • If k > 0: scales the vector's length by |k| without changing direction.
  • If k < 0: scales the length by |k| and reverses direction.
  • If k = 0: produces the zero vector.
• Example: 2⟨3, 4⟩ = ⟨6, 8⟩ (doubles length); (−2)⟨3, 4⟩ = ⟨−6, −8⟩ (doubles length, reverses direction).

🔢 Properties of scalar multiplication

Property	Statement
Associative	(kr)v⃗ = k(rv⃗)
Identity	1v⃗ = v⃗
Additive inverse	−v⃗ = (−1)v⃗
Distributive (scalar addition)	(k + r)v⃗ = kv⃗ + rv⃗
Distributive (vector addition)	k(v⃗ + w⃗) = kv⃗ + kw⃗
Zero product	kv⃗ = 0⃗ if and only if k = 0 or v⃗ = 0⃗

• These properties allow algebraic manipulation of vectors similar to variables.
• Example: solving 5v⃗ − 2(v⃗ + ⟨1, −2⟩) = 0⃗ for v⃗ uses distributive and associative properties step-by-step to isolate v⃗.

📏 Magnitude and direction

📏 Magnitude (length)

The magnitude of v⃗ = ⟨v₁, v₂⟩ is ‖v⃗‖ = √(v₁² + v₂²).

• Magnitude is always non-negative: ‖v⃗‖ ≥ 0.
• ‖v⃗‖ = 0 if and only if v⃗ = 0⃗.
• For scalar k: ‖kv⃗‖ = |k|‖v⃗‖ (absolute value of k times the magnitude).
• Example: ‖⟨3, 4⟩‖ = √(9 + 16) = 5.

🧭 Direction (unit vector)

If v⃗ ≠ 0⃗, the direction of v⃗ is the unit vector v̂ = ⟨cos(θ), sin(θ)⟩, where θ is the angle the vector makes with the positive x-axis (in standard position).

• A unit vector has magnitude 1.
• The direction v̂ can be found by normalizing: v̂ = (1/‖v⃗‖)v⃗.
• Any vector can be written as v⃗ = ‖v⃗‖v̂ (magnitude times direction).
• Example: for v⃗ = ⟨3, 4⟩ with ‖v⃗‖ = 5, the direction is v̂ = (1/5)⟨3, 4⟩ = ⟨3/5, 4/5⟩.

🔄 Converting between forms

• Polar to component: if ‖v⃗‖ = r and direction angle is θ, then v⃗ = ⟨r cos(θ), r sin(θ)⟩.
• Component to polar: given v⃗ = ⟨v₁, v₂⟩, find ‖v⃗‖ = √(v₁² + v₂²) and θ using tan(θ) = v₂/v₁ (adjust for quadrant).
• Resolving a vector into components: using magnitude and direction information to find component form.

🎯 Principal unit vectors

The principal unit vectors are î = ⟨1, 0⟩ (positive x-direction) and ĵ = ⟨0, 1⟩ (positive y-direction).

• Any vector v⃗ = ⟨v₁, v₂⟩ can be written as v⃗ = v₁î + v₂ĵ (principal vector decomposition).
• This expresses v⃗ as a combination of horizontal and vertical components.
• Example: ⟨3, 4⟩ = 3î + 4ĵ.

🌍 Applications: velocity and force

✈️ Velocity problems

• Velocity couples speed (magnitude) with direction.
• Airspeed vs ground speed: an airplane's velocity relative to air (v⃗) plus wind velocity (w⃗) gives true velocity (v⃗ + w⃗).
• Example: plane flies at 175 mph bearing N40°E; wind blows 35 mph bearing S60°E. Resolve both into components, add them, then find magnitude (true speed) and direction (true bearing) of the resultant.
• Don't confuse: bearing angles (measured from north/south toward east/west) vs standard position angles (measured counterclockwise from positive x-axis).

⚖️ Force and equilibrium

• A force is a push or pull with magnitude (measured in pounds or Newtons) and direction.
• Static equilibrium: for an object to remain stationary, the sum of all force vectors must be 0⃗.
• Example: a 50-pound speaker suspended by two cables at angles 60° and 30° with the ceiling. Let w⃗ = weight (pointing down), T⃗₁ and T⃗₂ = tensions in cables (pointing up at angles). Require w⃗ + T⃗₁ + T⃗₂ = 0⃗. Resolve each into components, equate x- and y-components to zero, solve the resulting system for ‖T⃗₁‖ and ‖T⃗₂‖.
• Key technique: write each force in component form using magnitude and direction, then set up a system of linear equations from the equilibrium condition.

🔍 Resolving vectors in applications

• Start with magnitude and direction information (e.g., speed and bearing, or force and angle).
• Convert to component form using v⃗ = ‖v⃗‖⟨cos(θ), sin(θ)⟩.
• Perform vector addition/subtraction component-wise.
• Convert result back to magnitude and direction if needed.
• Example workflow: plane velocity + wind velocity → resultant velocity → compute magnitude (true speed) and angle (true bearing).

🔧 Special concepts and techniques

🔧 Normalizing a vector

• Normalizing means converting a nonzero vector into a unit vector pointing in the same direction.
• Formula: v̂ = (1/‖v⃗‖)v⃗.
• Geometrically: shrink (or stretch) the vector's terminal point (in standard position) to lie on the unit circle.
• Every unit vector has magnitude 1; conversely, any nonzero vector can be normalized.

🔧 Parallel vectors

• Two nonzero vectors v⃗ and w⃗ are parallel if they have the same or opposite directions.
• Algebraically: v⃗ = kw⃗ for some nonzero scalar k.
  • If k > 0: same direction.
  • If k < 0: opposite directions.
• Don't confuse: parallel vectors can have different magnitudes; they differ only by a scalar multiple.

🔧 The Parallelogram Law

• For any vectors v⃗ and w⃗: ‖v⃗‖² + ‖w⃗‖² = (1/2)[‖v⃗ + w⃗‖² + ‖v⃗ − w⃗‖²].
• Relates the magnitudes of two vectors to the magnitudes of their sum and difference.
• Geometric interpretation: in the parallelogram formed by v⃗ and w⃗, the sum of the squares of the side lengths equals half the sum of the squares of the diagonal lengths.

🔧 Vectors and lines

• A non-vertical line y = mx + b can be described using vectors.
• Let v⃗₀ = ⟨0, b⟩ (position vector of the y-intercept) and s⃗ = ⟨1, m⟩ (slope vector).
• Any point on the line corresponds to the terminal point of v⃗ = v⃗₀ + ts⃗ for some scalar t.
• This expresses the line as a collection of vectors: a fixed starting point plus all scalar multiples of a direction vector.

🧭 Overview

🧠 One-sentence thesis

The dot product multiplies two vectors to produce a scalar that reveals the angle between them and measures how much of one vector lies in the direction of another, enabling decomposition of vectors into parallel and orthogonal components.

📌 Key points (3–5)

• What the dot product is: A binary operation that takes two vectors and produces a scalar (not another vector).
• Geometric meaning: The dot product encodes the angle between vectors via the formula v·w = ‖v‖‖w‖cos(θ).
• Orthogonality detection: Two nonzero vectors are perpendicular (orthogonal) if and only if their dot product equals zero.
• Common confusion: The dot product is also called the "scalar product" because it outputs a scalar, not because it involves multiplying by scalars.
• Projection application: The dot product enables finding the orthogonal projection of one vector onto another, decomposing any vector into parallel and perpendicular components.

🔢 Definition and computation

🔢 The dot product formula

Dot product: If v = ⟨v₁, v₂⟩ and w = ⟨w₁, w₂⟩, then v·w = v₁w₁ + v₂w₂.

• Multiply corresponding components, then add the results.
• The output is always a single number (scalar), never a vector.
• Example: ⟨3, 4⟩·⟨1, −2⟩ = (3)(1) + (4)(−2) = 3 − 8 = −5.

🔄 Algebraic properties

The dot product behaves like ordinary multiplication in many ways:

Property	Formula	What it means
Commutative	v·w = w·v	Order doesn't matter
Distributive	u·(v + w) = u·v + u·w	Distributes over addition
Scalar property	(kv)·w = k(v·w) = v·(kw)	Scalars factor out
Relation to magnitude	v·v = ‖v‖²	Dot product with itself gives squared length

• These properties follow from the definition and real number arithmetic.
• The excerpt proves commutative and scalar properties step-by-step using component forms.

📐 Geometric interpretation

📐 The angle formula

Geometric interpretation: For nonzero vectors v and w, v·w = ‖v‖‖w‖cos(θ), where θ is the angle between v and w (with 0 ≤ θ ≤ π).

• This connects the algebraic dot product to geometry.
• The angle θ is measured when both vectors start at the same point.
• Proof uses the Law of Cosines applied to the triangle formed by v, w, and v − w.

🧮 Finding angles between vectors

Solving the geometric formula for θ gives:

θ = arccos(v·w / (‖v‖‖w‖)) = arccos(v̂·ŵ)

• The second form uses unit vectors (normalized versions).
• Example: For v = ⟨3, −3√3⟩ and w = ⟨−√3, 1⟩, compute v·w = −6√3, ‖v‖ = 6, ‖w‖ = 2, so θ = arccos(−√3/2) = 5π/6 (or 150°).
• When the result isn't a common angle, leave the answer in arccos form.

⊥ Orthogonality

⊥ Detecting perpendicular vectors

Orthogonality test: Nonzero vectors v and w are orthogonal (written v ⊥ w) if and only if v·w = 0.

• Orthogonal means the angle between them is π/2 radians (90°).
• Since cos(π/2) = 0, the geometric formula v·w = ‖v‖‖w‖cos(θ) gives v·w = 0.
• Example: ⟨2, 2⟩·⟨5, −5⟩ = 10 − 10 = 0, so these vectors are orthogonal.

⊥ Application to perpendicular lines

The excerpt uses the dot product to prove a familiar fact about slopes:

• Lines y = m₁x + b₁ and y = m₂x + b₂ are perpendicular if and only if m₁m₂ = −1.
• Proof strategy: construct direction vectors ⟨1, m₁⟩ and ⟨1, m₂⟩ for each line, then show they are orthogonal exactly when 1 + m₁m₂ = 0.

Don't confuse: There is no "zero product property" for dot products—v·w can equal zero even when neither v nor w is the zero vector (they're just perpendicular).

🎯 Orthogonal projection

🎯 What projection means

Orthogonal projection: The orthogonal projection of v onto w, denoted proj_w(v), is the vector p that lies along w and satisfies: v = p + q where q is perpendicular to w.

• Geometrically: drop a perpendicular from the tip of v to the line containing w.
• The projection p points in the same direction as w (or opposite if the angle is obtuse).
• The formula is proj_w(v) = (v·ŵ)ŵ, where ŵ is the unit vector in the direction of w.

🎯 Alternative projection formulas

Three equivalent ways to compute projection:

proj_w(v) = (v·ŵ)ŵ = ((v·w)/‖w‖²)w = ((v·w)/(w·w))w

• The first form is conceptually clearest (scalar projection times direction).
• The last two forms avoid computing the unit vector ŵ explicitly, which can involve messy square roots.
• Example: For v = ⟨1, 8⟩ and w = ⟨−1, 2⟩, compute v·w = 15 and w·w = 5, so proj_w(v) = (15/5)⟨−1, 2⟩ = ⟨−3, 6⟩.

🎯 The scalar projection

The quantity v·ŵ is called the scalar projection of v onto w:

• It measures "how much of v is in the direction of w."
• It's the signed length of the projection vector.
• Positive when the angle is acute, negative when obtuse, zero when perpendicular.

🧩 Vector decomposition

🧩 Splitting into parallel and perpendicular parts

Generalized Decomposition Theorem: Any vector v can be uniquely written as v = p + q where p is parallel to w (i.e., p = kw for some scalar k) and q is orthogonal to w (i.e., q·w = 0).

• Take p = proj_w(v) and q = v − p.
• The excerpt proves q·w = 0 by direct computation using projection formulas.
• The proof also shows this decomposition is unique (only one way to split v this way).

🧩 Verifying orthogonality

To check that a projection is correct:

1. Verify p is a scalar multiple of w (correct direction).
2. Compute q = v − p.
3. Check that q·w = 0 (orthogonality condition).

Example: With v = ⟨1, 8⟩, p = ⟨−3, 6⟩, and w = ⟨−1, 2⟩, we get q = ⟨4, 2⟩ and q·w = −4 + 4 = 0 ✓.

⚙️ Application to work

⚙️ Work done by a force

Work as a dot product: When a constant force F is applied to move an object along the displacement vector PQ, the work done is W = F·PQ = ‖F‖‖PQ‖cos(θ), where θ is the angle between the force and the motion.

• Only the component of force in the direction of motion does work.
• This is exactly what the dot product F·P̂Q measures.
• Units: if force is in pounds and distance in feet, work is in foot-pounds.

⚙️ Computing work in practice

Two equivalent approaches:

1. Vector method: Find component forms of F and PQ, then compute F·PQ directly.
2. Magnitude-angle method: Use W = ‖F‖‖PQ‖cos(θ) with known magnitudes and angle.

Example: Taylor pulls a wagon with 10 pounds of force at a 30° angle for 50 feet. Using the magnitude-angle formula: W = (10)(50)cos(30°) = 250√3 foot-pounds.

Don't confuse: The angle θ in the work formula is between the force vector and the displacement vector, not necessarily the angle mentioned in the problem setup (e.g., the incline angle).

🧭 Overview

🧠 One-sentence thesis

Parametric equations allow us to describe curves that fail both the Vertical and Horizontal Line Tests by expressing both x and y as separate functions of a third variable (the parameter t), thereby enabling us to study a much wider variety of curves and to capture their orientation.

📌 Key points (3–5)

• What parametric equations are: a system {x = f(t), y = g(t)} where both coordinates are functions of a parameter t, allowing us to trace curves that cannot be expressed as y = f(x) or x = g(y).
• Orientation matters: the parametrization determines a direction of motion along the curve (indicated by arrows); the curve itself has no inherent orientation, but the parametrization imposes one.
• Eliminating the parameter: we can often convert parametric equations back to a single equation in x and y using substitution or trigonometric identities, though this loses orientation information.
• Common confusion: the same curve can have infinitely many different parametrizations; changing the parameter (e.g., replacing t with –t or t – c) alters orientation or timing but may trace the same geometric path.
• Standard formulas exist: lines, circles, ellipses, and parabolas all have standard parametric forms that can be adjusted for orientation and starting point.

🐛 Motivation and core concept

🐛 Why parametric equations?

• Many interesting curves cannot be written as y = f(x) because they fail the Vertical Line Test.
• Similarly, they may fail the Horizontal Line Test, so x = g(y) also doesn't work.
• Solution: imagine a bug crawling along the curve. At any time t, the bug is at exactly one point P(x, y). We can describe x as a function of t and y as a (usually different) function of t.
• Example: the bug starts at point O, loops around, crosses its own path at point Q (at two different times), and continues—this motion is naturally described by parametric equations.

📐 Definition

Parametric equations (parametrization): a system of equations {x = f(t), y = g(t)} where the independent variable t is called the parameter. The curve C traced out is the set of all points (f(t), g(t)) as t varies over some interval.

• The parameter t often represents time, but it can be any variable.
• The parametrization endows the curve with an orientation (direction of motion) indicated by arrows showing increasing t.
• The curve itself (as a set of points) has no orientation; the parametrization provides it.

🔄 Familiar example: circular motion

• The Unit Circle is parametrized by {x = cos(t), y = sin(t)} for 0 ≤ t < 2π, giving counter-clockwise orientation.
• More generally, {x = r cos(ωt), y = r sin(ωt)} traces a circle of radius r centered at the origin.
  • If ω > 0: counter-clockwise.
  • If ω < 0: clockwise.
  • The angular frequency ω determines "how fast" the motion is.

📊 Sketching parametric curves

📊 General procedure

1. Choose friendly values of t and compute corresponding (x(t), y(t)) points.
2. Plot the points in order of increasing t.
3. Draw arrows to indicate the direction (orientation) of motion.
4. Analyze behavior: look at how x(t) and y(t) individually increase or decrease to understand the path.
5. Eliminate the parameter (when possible) to identify the underlying curve type (parabola, ellipse, hyperbola, etc.).

🧮 Example: parabola from parametric form

• Given {x = t² – 3, y = 2t – 1} for t ≥ –2.
• Plot points: t = –2 gives (1, –5); t = –1 gives (–2, –3); t = 0 gives (–3, –1); t = 1 gives (–2, 1); t = 2 gives (1, 3); t = 3 gives (6, 5).
• Eliminate parameter: solve y = 2t – 1 for t to get t = (y + 1)/2, substitute into x = t² – 3 to obtain (y + 1)² = 4(x + 3), a parabola opening to the right with vertex (–3, –1).
• The parametric form traces only the portion where t ≥ –2, corresponding to y ≥ –5.
• Key point: eliminating the parameter identifies the curve type but loses orientation information.

🔍 Analyzing motion

• Look at whether x(t) and y(t) are increasing or decreasing over intervals.
• Example: if x(t) is increasing and y(t) is decreasing, the motion is to the right and downward.
• Check endpoints and any critical points (where derivatives would be zero in calculus).

🔄 Eliminating the parameter

🔄 Algebraic functions

• Use substitution or elimination techniques from systems of equations.
• Solve one equation for t, substitute into the other.
• Example: {x = t³, y = 2t²} for –1 ≤ t ≤ 1. Solve x = t³ for t to get t = ∛x, substitute into y = 2t² to get y = 2(∛x)² = 2x^(2/3).

🔄 Exponential functions

• Example: {x = 2e^(–t), y = e^(–2t)} for t ≥ 0.
• Note that y = e^(–2t) = (e^(–t))². Since x = 2e^(–t) means e^(–t) = x/2, we get y = (x/2)² = x²/4.
• The parametric form traces only part of the parabola y = x²/4 (starting at (2, 1) as t = 0 and approaching (0, 0) as t → ∞).

🔄 Trigonometric functions

• Use trigonometric identities, especially the Pythagorean Identity cos²(t) + sin²(t) = 1.
• Example: {x = sin(t), y = csc(t)} for 0 < t < π.
  • Since y = csc(t) = 1/sin(t) and x = sin(t), we get y = 1/x.
  • The parametric form traces part of the hyperbola y = 1/x twice as t runs from 0 to π.
• Example: {x = 1 + 3cos(t), y = 2sin(t)} for 0 ≤ t ≤ 3π/2.
  • Solve for cos(t) = (x – 1)/3 and sin(t) = y/2.
  • Substitute into cos²(t) + sin²(t) = 1 to get ((x – 1)/3)² + (y/2)² = 1, or (x – 1)²/9 + y²/4 = 1, an ellipse centered at (1, 0).
  • The parametric form traces three-quarters of the ellipse (counter-clockwise).

⚠️ Don't confuse

• Eliminating the parameter gives the equation of the curve but loses orientation.
• The parametric form may trace only part of the curve described by the Cartesian equation.
• Restrictions on t translate to restrictions on x and y, but these can be tricky to express after elimination.

📐 Standard parametrizations

📐 Common curves formulas

Curve type	Parametrization	Notes
y = f(x) for x in interval I	{x = t, y = f(t)}, t in I	Simplest case
x = g(y) for y in interval I	{x = g(t), y = t}, t in I	Swap roles
Line segment from (x₀, y₀) to (x₁, y₁)	{x = x₀ + (x₁ – x₀)t, y = y₀ + (y₁ – y₀)t}, 0 ≤ t ≤ 1	"Start + displacement·t"
Circle/ellipse (x–h)²/a² + (y–k)²/b² = 1	{x = h + a cos(t), y = k + b sin(t)}, 0 ≤ t < 2π	Counter-clockwise

📐 Line segments

• Formula: {x = x₀ + (x₁ – x₀)t, y = y₀ + (y₁ – y₀)t} for 0 ≤ t ≤ 1.
• Think: "starting point + (displacement)t".
• Example: segment from (2, –3) to (1, 5).
  • Displacement in x: 1 – 2 = –1, so x = 2 – t.
  • Displacement in y: 5 – (–3) = 8, so y = –3 + 8t.
  • Result: {x = 2 – t, y = –3 + 8t} for 0 ≤ t ≤ 1.

📐 Circles and ellipses

• For a circle (x – h)² + (y – k)² = r², use {x = h + r cos(t), y = k + r sin(t)} for 0 ≤ t < 2π (counter-clockwise).
• For an ellipse (x – h)²/a² + (y – k)²/b² = 1, use {x = h + a cos(t), y = k + b sin(t)} for 0 ≤ t < 2π.
• These come from the Pythagorean Identity: if cos(t) = (x – h)/a and sin(t) = (y – k)/b, then cos²(t) + sin²(t) = 1 gives the ellipse equation.

📐 Parabolas

• For y = x², use {x = t, y = t²}.
• For x = y², use {x = t², y = t}.
• Adjust bounds on t to match the desired portion of the parabola.

🔧 Adjusting parametrizations

🔧 Reversing orientation

Reversing orientation: Replace every occurrence of t with –t (including in the bounds on t) to reverse the direction of motion along the curve.

• Example: {x = t, y = t²} for –1 ≤ t ≤ 2 goes from (–1, 1) to (2, 4).
• Replace t with –t: {x = –t, y = (–t)² = t²} for –1 ≤ –t ≤ 2, which simplifies to {x = –t, y = t²} for –2 ≤ t ≤ 1.
• This now goes from (2, 4) to (–1, 1), the opposite direction.

🔧 Shifting the parameter

Shift of parameter: Replace every occurrence of t with (t – c) (including in the bounds) to introduce a "time delay" of c units, shifting the start of the parameter ahead by c.

• Example: {x = 3t, y = 4t} for 0 ≤ t ≤ 1 describes a line segment from (0, 0) to (3, 4).
• To shift so it starts at t = 1 instead of t = 0, replace t with (t – 1): {x = 3(t – 1), y = 4(t – 1)} for 0 ≤ t – 1 ≤ 1.
• Simplify: {x = 3t – 3, y = 4t – 4} for 1 ≤ t ≤ 2.
• Now the segment is traced as t runs from 1 to 2 instead of 0 to 1.

🔧 Combining adjustments

• You can reverse orientation and then shift the parameter (or vice versa) to get the desired behavior.
• Example: parametrize the Unit Circle clockwise starting at (0, –1) when t = 0.
  • Start with {x = cos(t), y = sin(t)} (counter-clockwise, starts at (1, 0) when t = 0).
  • Reverse: replace t with –t to get {x = cos(–t) = cos(t), y = sin(–t) = –sin(t)} for –2π < t ≤ 0 (clockwise, still starts at (1, 0) when t = 0).
  • Shift to start at (0, –1): the point (0, –1) occurs at t = –3π/2, so shift by 3π/2 by replacing t with (t – 3π/2).
  • Result: {x = cos(t – 3π/2) = –sin(t), y = –sin(t – 3π/2) = –cos(t)} for 0 ≤ t < 2π.

🌀 Special curve: the cycloid

🌀 Definition and derivation

• A cycloid is the path traced by a point on the rim of a circle of radius r as the circle rolls along a straight line (e.g., the positive x-axis) at constant velocity v.
• Let θ be the angle (in radians) measuring the amount of clockwise rotation.
• The point starts at the bottom of the circle and traces out arches as the circle rolls.

🌀 Parametric equations

• From the excerpt's derivation: {x = r(θ – sin(θ)), y = r(1 – cos(θ))} for θ ≥ 0.
• Here θ is the parameter (often renamed t).
• One complete arch corresponds to 0 ≤ θ < 2π (one full revolution of the circle).
• The center of the rolling circle is at (rθ, r) at parameter value θ.

🌀 Graphing with technology

• Example: for r = 3, the equations are {x = 3(t – sin(t)), y = 3(1 – cos(t))} for t ≥ 0.
• One arch: 0 ≤ t < 2π, so x ranges from 0 to 6π, y ranges from 0 to 6.
• Three arches: 0 ≤ t < 6π, so x ranges from 0 to 18π.
• Use calculator in Parametric Mode and radian mode; adjust window settings to see the true geometry (may need Zoom Square).

🎯 Applications

🎯 Projectile motion

• An object launched into the air (ignoring air resistance) follows parametric equations:
  • {x = v₀ cos(θ) t, y = –(1/2)gt² + v₀ sin(θ) t + s₀} for 0 ≤ t ≤ T.
  • v₀ = initial speed, θ = launch angle from horizontal, g = acceleration due to gravity, s₀ = initial height, T = time when object returns to ground.
• Eliminating the parameter gives a parabolic path: y = –(g sec²(θ))/(2v₀²) x² + tan(θ) x + s₀.
• Maximum height occurs at x = (v₀² sin(2θ))/(2g), with height y = (v₀² sin²(θ))/(2g) + s₀.

🎯 Multi-part paths

• Parametric equations can describe paths with multiple segments by using piecewise-defined functions.
• Example: a path that goes from (0, 0) to (3, 4) along a line, then from (3, 4) to (5, 0) along another line.
  • First segment: {x = 3t, y = 4t} for 0 ≤ t ≤ 1.
  • Second segment (shifted to start at t = 1): {x = 1 + 2t, y = 8 – 4t} for 1 ≤ t ≤ 2.
  • Combined: {x = f(t), y = g(t)} for 0 ≤ t ≤ 2, where f and g are piecewise-defined.

🎯 Polar curves as parametric equations

• Any polar curve r = f(θ) can be converted to parametric form: {x = f(θ) cos(θ), y = f(θ) sin(θ)}.
• This allows graphing polar curves using parametric mode on a calculator.

⚠️ Common pitfalls

⚠️ Orientation is not intrinsic

• The curve (as a set of points) has no orientation.
• Different parametrizations of the same curve can give different orientations.
• Eliminating the parameter loses all orientation information.

⚠️ Parametric form may trace only part of a curve

• Example: {x = sin(t), y = csc(t)} for 0 < t < π traces part of y = 1/x twice, not the entire hyperbola.
• After eliminating the parameter, you must carefully determine which portion of the Cartesian curve corresponds to the parametric form.

⚠️ Infinitely many parametrizations

• Any given curve has infinitely many different parametrizations.
• Adjusting the parameter (reversing, shifting, or using a different function of t) gives a new parametrization.
• Standard formulas are just one convenient choice among many.

⚠️ Bounds on t matter

• The interval for t determines which portion of the curve is traced.
• Changing the bounds changes the curve segment, even if the equations for x(t) and y(t) stay the same.