AP Precalculus · Unit 3

Trigonometric
& Polar Functions

From the unit circle to the polar plane — a complete visual study guide you can work through at your own pace.

I II III IV 1 i cos θ sin θ θ
3.1

Radian Measure & the Unit Circle

Radians are the natural way to measure angles — they connect arc length directly to radius, making calculus much cleaner than degrees ever could.

Core Idea

What is a Radian?

One radian is the angle created when the arc length equals the radius. Since a full circle has circumference $2\pi r$, a full rotation is exactly $2\pi$ radians.

$$\theta = \frac{s}{r}$$
arc length $s$, radius $r$
Conversion

Degrees ↔ Radians

Multiply by the right conversion factor. Always simplify the fraction — don't leave ugly decimals.

$$\text{deg} \times \frac{\pi}{180} = \text{rad}$$ $$\text{rad} \times \frac{180}{\pi} = \text{deg}$$
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Memorize these benchmark angles:
$0°=0$, $\;30°=\dfrac{\pi}{6}$, $\;45°=\dfrac{\pi}{4}$, $\;60°=\dfrac{\pi}{3}$, $\;90°=\dfrac{\pi}{2}$, $\;180°=\pi$, $\;270°=\dfrac{3\pi}{2}$, $\;360°=2\pi$

Key Unit Circle Coordinates

Angle (deg)Angle (rad)cos θ (x)sin θ (y)tan θ
0°$0$$1$$0$$0$
30°$\frac{\pi}{6}$$\frac{\sqrt{3}}{2}$$\frac{1}{2}$$\frac{\sqrt{3}}{3}$
45°$\frac{\pi}{4}$$\frac{\sqrt{2}}{2}$$\frac{\sqrt{2}}{2}$$1$
60°$\frac{\pi}{3}$$\frac{1}{2}$$\frac{\sqrt{3}}{2}$$\sqrt{3}$
90°$\frac{\pi}{2}$$0$$1$undefined
180°$\pi$$-1$$0$$0$
270°$\frac{3\pi}{2}$$0$$-1$undefined
Worked Example Converting and Finding Exact Values
Problem: Convert $240°$ to radians, then find $\sin(240°)$, $\cos(240°)$, and $\tan(240°)$ exactly.
  1. Convert to radians: $240° \times \dfrac{\pi}{180} = \dfrac{240\pi}{180} = \dfrac{4\pi}{3}$
  2. Locate the angle: $\dfrac{4\pi}{3}$ is in Quadrant III ($\pi < \theta < \dfrac{3\pi}{2}$). Its reference angle is $\dfrac{4\pi}{3} - \pi = \dfrac{\pi}{3}$ (which is $60°$).
  3. Apply ASTC rule: In Q III, both sin and cos are negative, tan is positive.
  4. Write the answers: $\sin\!\left(\dfrac{4\pi}{3}\right) = -\dfrac{\sqrt{3}}{2}$, $\;\cos\!\left(\dfrac{4\pi}{3}\right) = -\dfrac{1}{2}$, $\;\tan\!\left(\dfrac{4\pi}{3}\right) = \sqrt{3}$
✓   Reference angle + ASTC sign rule = exact values without a calculator
3.2

Trigonometric Functions

Six functions, all born from the unit circle. Understand their definitions and graphs, and everything else follows naturally.

Primary Trio

sin, cos, tan

For a point $(x, y)$ on the unit circle at angle $\theta$:

$$\sin\theta = y \qquad \cos\theta = x \qquad \tan\theta = \frac{y}{x}$$

On the unit circle, the radius is 1, so the coordinates are the trig values directly.

Reciprocal Trio

csc, sec, cot

Each is simply 1 divided by a primary function:

$$\csc\theta = \frac{1}{\sin\theta} \quad \sec\theta = \frac{1}{\cos\theta} \quad \cot\theta = \frac{1}{\tan\theta}$$
🧭
ASTC — "All Students Take Calculus"
Q I: All positive · Q II: Sine positive · Q III: Tangent positive · Q IV: Cosine positive

Graph Gallery

Sine · $y = \sin\theta$

Key Features

• Domain: all reals  • Range: $[-1,\,1]$
• Period: $2\pi$  • Odd function: $\sin(-x)=-\sin(x)$
• Zeros at $n\pi$, peaks at $\frac{\pi}{2}+2n\pi$

Cosine · $y = \cos\theta$

Key Features

• Domain: all reals  • Range: $[-1,\,1]$
• Period: $2\pi$  • Even function: $\cos(-x)=\cos(x)$
• Zeros at $\frac{\pi}{2}+n\pi$, peaks at $2n\pi$

Worked Example Evaluating Without a Calculator
Problem: Find the exact value of $\tan\!\left(\dfrac{7\pi}{6}\right)$.
  1. Reference angle: $\dfrac{7\pi}{6} - \pi = \dfrac{\pi}{6}$, so the reference angle is $30°$.
  2. Quadrant: $\pi < \dfrac{7\pi}{6} < \dfrac{3\pi}{2}$ → Quadrant III → tan is positive in Q III.
  3. Reference value: $\tan\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$
  4. Apply sign: $\tan\!\left(\dfrac{7\pi}{6}\right) = +\dfrac{\sqrt{3}}{3}$
✓   $\tan\!\left(\dfrac{7\pi}{6}\right) = \dfrac{\sqrt{3}}{3}$
3.3

Amplitude, Period, Phase & Vertical Shift

One master formula controls four independent transformations. Identify each parameter and you can describe or sketch any sinusoidal function.

📐
The General Sinusoidal Form: $$y = A\sin(B(x - C)) + D \qquad \text{or} \qquad y = A\cos(B(x - C)) + D$$
$|A|$

Amplitude

The height from midline to peak. Range becomes $[D-|A|,\; D+|A|]$. If $A<0$, the graph flips vertically.

$B$

Period

Controls how fast the wave repeats.

$$\text{Period} = \frac{2\pi}{|B|}$$
$C$

Phase Shift

Horizontal translation. Positive $C$ → right, negative $C$ → left.

$$\text{Shift} = C$$
$D$

Vertical Shift

Moves the midline up or down. The midline is the horizontal axis of symmetry of the wave.

Worked Example Identify All Parameters
Problem: For $y = -3\sin\!\left(2x - \dfrac{\pi}{2}\right) + 1$, identify the amplitude, period, phase shift, vertical shift, and state the range.
  1. Rewrite in standard form: Factor out $B=2$ from the argument: $y = -3\sin\!\left(2\!\left(x - \dfrac{\pi}{4}\right)\!\right) + 1$
  2. Amplitude: $|A| = |-3| = 3$. The negative sign means the graph is flipped (starts going down).
  3. Period: $\dfrac{2\pi}{|B|} = \dfrac{2\pi}{2} = \pi$
  4. Phase shift: $C = \dfrac{\pi}{4}$ to the right.
  5. Vertical shift: $D = 1$ up; the midline is $y = 1$.
  6. Range: $[D - |A|,\; D + |A|] = [1-3,\;1+3] = [-2,\;4]$
✓   Amplitude 3, Period $\pi$, Phase shift $\dfrac{\pi}{4}$ right, Midline $y=1$, Range $[-2,\,4]$
Worked Example Write an Equation from a Graph
Problem: A sinusoidal function has a maximum of $7$ at $x = \dfrac{\pi}{6}$, a minimum of $1$, and a period of $\dfrac{2\pi}{3}$. Write an equation.
  1. Amplitude: $A = \dfrac{\text{max}-\text{min}}{2} = \dfrac{7-1}{2} = 3$
  2. Midline / Vertical shift: $D = \dfrac{\text{max}+\text{min}}{2} = \dfrac{7+1}{2} = 4$
  3. Find $B$: Period $= \dfrac{2\pi}{3}$ so $B = \dfrac{2\pi}{2\pi/3} = 3$
  4. Phase shift: A cosine peaks at its phase shift. Since max is at $x = \dfrac{\pi}{6}$: $C = \dfrac{\pi}{6}$
  5. Equation: $y = 3\cos\!\left(3\!\left(x - \dfrac{\pi}{6}\right)\!\right) + 4$
✓   $y = 3\cos\!\left(3x - \dfrac{\pi}{2}\right) + 4$
3.4

Inverse Trigonometric Functions

Because trig functions repeat, we must restrict their domains to create true inverses. These restricted ranges are non-negotiable — you must memorize them.

Definition

What $\arcsin(x)$ means

$\arcsin(x)$ answers: "What angle has a sine of $x$?" — but only within the restricted range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

$$y = \arcsin(x) \iff \sin(y) = x$$
Key Restriction

Why We Restrict

$\sin(\pi/6) = 0.5$ and $\sin(5\pi/6) = 0.5$ — a function can't give two outputs for one input. We restrict so each inverse is a proper function.

FunctionDomainRange (Restricted)Quadrants
$y = \arcsin(x)$$[-1,\,1]$$\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right]$Q I & IV
$y = \arccos(x)$$[-1,\,1]$$[0,\,\pi]$Q I & II
$y = \arctan(x)$all reals$\left(-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right)$Q I & IV
⚠️
Common mistake: $\arcsin(\sin(\theta)) = \theta$ is only true when $\theta$ is already in the restricted range $[-\pi/2,\,\pi/2]$. For example: $\arcsin(\sin(2\pi/3)) \ne 2\pi/3$ — it equals $\pi/3$.
Worked Example Exact Inverse Values
Problem: Find the exact value of $\cos\!\left(\arctan\!\left(-\sqrt{3}\right)\right)$.
  1. Inner function: Let $\theta = \arctan(-\sqrt{3})$. We need $\tan\theta = -\sqrt{3}$ with $\theta\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$.
  2. Reference: $\tan(\pi/3)=\sqrt{3}$, so with the negative, $\theta = -\dfrac{\pi}{3}$.
  3. Outer function: $\cos\!\left(-\dfrac{\pi}{3}\right) = \cos\!\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$ (cosine is even).
✓   $\cos\!\left(\arctan\!\left(-\sqrt{3}\right)\right) = \dfrac{1}{2}$
3.5

Trigonometric Identities

Identities are equations true for all valid angles. They're your algebraic toolkit for simplifying expressions, solving equations, and proving results.

Essential Identity Reference

Pythagorean
$$\sin^2\theta + \cos^2\theta = 1$$
Pythagorean Variant 1
$$1 + \tan^2\theta = \sec^2\theta$$
Pythagorean Variant 2
$$1 + \cot^2\theta = \csc^2\theta$$
Sum (sine)
$$\sin(A\pm B)$$$$= \sin A\cos B \pm \cos A\sin B$$
Sum (cosine)
$$\cos(A\pm B)$$$$= \cos A\cos B \mp \sin A\sin B$$
Double Angle (sin)
$$\sin 2\theta = 2\sin\theta\cos\theta$$
Double Angle (cos)
$$\cos 2\theta = \cos^2\!\theta - \sin^2\!\theta$$
Power Reduction
$$\sin^2\theta = \frac{1-\cos 2\theta}{2}$$
Quotient
$$\tan\theta = \frac{\sin\theta}{\cos\theta}$$
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Proving identities: Work on ONE side only. Start with the more complicated side. Convert everything to sin and cos if you're stuck. Never move terms across the equals sign.
Worked Example Prove an Identity
Problem: Prove that $\dfrac{\sin\theta}{1-\cos\theta} = \dfrac{1+\cos\theta}{\sin\theta}$.
  1. Work on the left side. Multiply numerator and denominator by $(1+\cos\theta)$:
  2. $$\frac{\sin\theta}{1-\cos\theta} \cdot \frac{1+\cos\theta}{1+\cos\theta} = \frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}$$
  3. Apply Pythagorean identity: $1-\cos^2\theta = \sin^2\theta$, so:
  4. $$\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta} = \frac{1+\cos\theta}{\sin\theta} \checkmark$$
✓   LHS = RHS. Identity proven.
Worked Example Solving a Trig Equation
Problem: Solve $2\cos^2 x - \cos x - 1 = 0$ for $x \in [0,\, 2\pi)$.
  1. Factor like a quadratic: Let $u = \cos x$. Then $2u^2 - u - 1 = (2u + 1)(u - 1) = 0$.
  2. Case 1: $u = 1 \Rightarrow \cos x = 1 \Rightarrow x = 0$
  3. Case 2: $2u + 1 = 0 \Rightarrow \cos x = -\dfrac{1}{2}$. Reference angle: $\dfrac{\pi}{3}$. Cosine is negative in Q II and Q III.
  4. $x = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$ and $x = \pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$
✓   $x = 0,\;\dfrac{2\pi}{3},\;\dfrac{4\pi}{3}$
3.6

Polar Coordinates & Graphs

Instead of $(x,y)$, a polar point is described by its distance from the origin and the angle it makes. Some curves that are messy in rectangular form become beautifully simple in polar.

Polar Coordinates

The $(r, \theta)$ System

$r$ = distance from the pole (origin), $\theta$ = angle from the polar axis (positive $x$-axis). Negative $r$ means go in the opposite direction.

$$\text{Point: }\;(r,\,\theta)$$
e.g. $(3,\,\pi/4)$ → distance 3 at 45°
Conversion Formulas

Polar ↔ Rectangular

$$x = r\cos\theta \qquad y = r\sin\theta$$ $$r^2 = x^2 + y^2 \qquad \tan\theta = \frac{y}{x}$$

Classic Polar Curves

Circle

$r = a$

A circle centered at the pole with radius $|a|$. Simple and perfect.

Rose Curve

$r = a\cos(n\theta)$

Petals! If $n$ is odd → $n$ petals. If $n$ is even → $2n$ petals. Traced as $\theta$ goes from $0$ to $2\pi$.

Limaçon

$r = a + b\cos\theta$

Shape depends on $|a/b|$: inner loop, cardioid ($a=b$), dimple, or convex. Cardioid is most common on exams.

Lemniscate

$r^2 = a^2\cos(2\theta)$

A figure-eight centered at the pole. Only exists where $\cos(2\theta) \geq 0$.

Left: $r = 2 + 2\cos\theta$ (cardioid) · Right: $r = \cos(3\theta)$ (3-petal rose)

Worked Example Converting Between Systems
Problem: Convert $r = 4\cos\theta$ to rectangular form and identify the curve.
  1. Multiply both sides by $r$: $r^2 = 4r\cos\theta$
  2. Substitute: $r^2 = x^2+y^2$ and $r\cos\theta = x$, giving $x^2 + y^2 = 4x$
  3. Complete the square: $(x-2)^2 + y^2 = 4$
  4. Identify: A circle centered at $(2, 0)$ with radius $2$.
✓   Circle: center $(2,\,0)$, radius $2$. In polar: $r = 4\cos\theta$
Worked Example Graphing a Cardioid
Problem: Sketch $r = 3 + 3\sin\theta$ and find its maximum distance from the pole.
  1. Identify type: $a = b = 3$, so this is a cardioid (heart-shaped). It's symmetric about the $y$-axis (because of $\sin\theta$).
  2. Key values:
    $\theta=0$: $r=3+0=3$
    $\theta=\pi/2$: $r=3+3=6$ ← maximum
    $\theta=\pi$: $r=3+0=3$
    $\theta=3\pi/2$: $r=3-3=0$ ← passes through pole
  3. Maximum $r$: occurs at $\theta=\pi/2$, giving $r_{\max} = 6$.
✓   Cardioid symmetric about $y$-axis; max distance from pole $= 6$ at $\theta = \pi/2$
Quick Review

Unit 3 Cheat Sheet

Must Know

Radian Conversions

$\times \frac{\pi}{180}$ for deg→rad · $\times \frac{180}{\pi}$ for rad→deg · Memorize the 5 benchmark angles

Must Know

ASTC in All Quadrants

Q I: all + · Q II: sin + · Q III: tan + · Q IV: cos +. Use reference angle for magnitude.

Must Know

Transformation Formula

$y=A\sin(B(x-C))+D$ · Period $=2\pi/B$ · Amplitude $=|A|$ · Always factor $B$ first.

Must Know

Inverse Ranges

arcsin: $[-\pi/2,\,\pi/2]$ · arccos: $[0,\,\pi]$ · arctan: $(-\pi/2,\,\pi/2)$

Must Know

Pythagorean Identities

$\sin^2+\cos^2=1$ is the root. Divide by $\cos^2$ for $1+\tan^2=\sec^2$. Divide by $\sin^2$ for $\cot^2+1=\csc^2$.

Must Know

Polar Conversions

$x=r\cos\theta$, $y=r\sin\theta$, $r^2=x^2+y^2$. Multiply by $r$ to use $r^2$ substitution.