Pre-Calculus Trigonometry

Unit 1 — Angle Measurement

U1Angles & Radian Measure

Angles measured in degrees or radians. A full rotation = $360°= 2\pi$ rad.

🔑 Key Conversion

$$\text{rad} = \text{deg} \times \frac{\pi}{180} \qquad \text{deg} = \text{rad} \times \frac{180}{\pi}$$ Arc length: $s = r\theta$ (θ in radians) | Sector area: $A = \tfrac{1}{2}r^2\theta$

📌 Must Memorize

$30°=\tfrac{\pi}{6}$, $45°=\tfrac{\pi}{4}$, $60°=\tfrac{\pi}{3}$, $90°=\tfrac{\pi}{2}$, $180°=\pi$, $270°=\tfrac{3\pi}{2}$, $360°=2\pi$

Worked Example

Convert $150°$ to radians.
$150 \times \dfrac{\pi}{180} = \dfrac{5\pi}{6}$

Answer: $\dfrac{5\pi}{6}$ rad

Unit 2 — The Unit Circle

U2Unit Circle & Special Angles

On the unit circle ($r=1$), each angle $\theta$ maps to point $(\cos\theta, \sin\theta)$.

🔑 Special Values

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$
$0$	$0$	$1$	$0$
$\pi/6$	$\tfrac{1}{2}$	$\tfrac{\sqrt3}{2}$	$\tfrac{1}{\sqrt3}$
$\pi/4$	$\tfrac{\sqrt2}{2}$	$\tfrac{\sqrt2}{2}$	$1$
$\pi/3$	$\tfrac{\sqrt3}{2}$	$\tfrac{1}{2}$	$\sqrt3$
$\pi/2$	$1$	$0$	undef

📌 ASTC Rule (Signs by Quadrant)

Q1: All + Q2: Sin + Q3: Tan + Q4: Cos +
Mnemonic: "All Students Take Calculus"

Unit 3 — Trig Identities

U3Fundamental & Pythagorean Identities

🔑 Must-Know Identities

Pythagorean:
$\sin^2\theta + \cos^2\theta = 1$
$1 + \tan^2\theta = \sec^2\theta$
$1 + \cot^2\theta = \csc^2\theta$

Reciprocal: $\csc\theta = \tfrac{1}{\sin\theta}$, $\sec\theta = \tfrac{1}{\cos\theta}$, $\cot\theta = \tfrac{1}{\tan\theta}$

Even/Odd: $\cos(-\theta)=\cos\theta$; $\sin(-\theta)=-\sin\theta$

🔑 Sum & Difference / Double Angle

$\sin(A\pm B)=\sin A\cos B \pm \cos A\sin B$
$\cos(A\pm B)=\cos A\cos B \mp \sin A\sin B$
$\sin 2\theta = 2\sin\theta\cos\theta$
$\cos 2\theta = \cos^2\theta - \sin^2\theta = 1-2\sin^2\theta = 2\cos^2\theta-1$

Worked Example

Simplify $\dfrac{\sin^2 x}{\cos^2 x} + \dfrac{1}{\sec^2 x}$
$= \tan^2 x + \cos^2 x$ ... use $\sec^2 x = 1+\tan^2 x$: this does not simplify further without more context — but $\tan^2 x \cdot \cos^2 x = \sin^2 x$.
Simpler example: Verify $\dfrac{1-\cos^2\theta}{\sin\theta} = \sin\theta$
$= \dfrac{\sin^2\theta}{\sin\theta} = \sin\theta$ ✓

Used: $\sin^2\theta + \cos^2\theta = 1$

Unit 4 — Graphs of Trig Functions

U4Graphing & Transformations

For $y = A\sin(Bx - C) + D$:

🔑 Key Parameters

Amplitude $= |A|$ · Period $= \dfrac{2\pi}{|B|}$
Phase shift $= \dfrac{C}{B}$ (right if positive)
Vertical shift $= D$

📌 Period of tan/cot

Period of $\tan$ and $\cot$ is $\dfrac{\pi}{|B|}$ (not $2\pi$).

Worked Example

Find the period of $y = 3\cos(2x - \pi) + 1$
$B=2$, Period $= \dfrac{2\pi}{2} = \pi$. Phase shift $= \dfrac{\pi}{2}$ (right). Amplitude $= 3$. Vertical shift $= +1$.

Period = $\pi$, Amplitude = 3

Unit 5 — Inverse Trig Functions

U5Inverse Trigonometric Functions

🔑 Ranges (Principal Values)

$\arcsin$: domain $[-1,1]$, range $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$
$\arccos$: domain $[-1,1]$, range $[0, \pi]$
$\arctan$: domain $\mathbb{R}$, range $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$

Worked Example

Evaluate $\arctan(-1)$
Need angle in $(-\pi/2, \pi/2)$ where $\tan\theta = -1$. Answer: $\theta = -\pi/4$.

$\arctan(-1) = -\dfrac{\pi}{4}$

Unit 6 — Law of Sines & Cosines

U6Law of Sines & Law of Cosines

🔑 Law of Sines

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ Use when: AAS, ASA, or ambiguous SSA case.

🔑 Law of Cosines

$c^2 = a^2 + b^2 - 2ab\cos C$
Use when: SAS or SSS.

Worked Example

Triangle: $a=7, b=10, C=60°$. Find $c$.
$c^2 = 49 + 100 - 2(7)(10)\cos60° = 149 - 70 = 79$
$c = \sqrt{79} \approx 8.888$

$c \approx 8.9$ (3 s.f.)

Unit 7 — Trig Equations & Solving

U7Solving Trigonometric Equations

Find all solutions in $[0, 2\pi)$ unless specified otherwise.

🔑 General Solutions

$\sin\theta = k$: ref angle $\alpha = \arcsin|k|$; solutions in Q1&Q2 if $k>0$, Q3&Q4 if $k<0$
$\cos\theta = k$: ref angle in Q1&Q4 if $k>0$, Q2&Q3 if $k<0$
General: $\sin\theta = k \Rightarrow \theta = \alpha + 2\pi n$ or $\theta = \pi - \alpha + 2\pi n$

Worked Example

Solve $2\sin x - \sqrt{3} = 0$ for $x \in [0, 2\pi)$
$\sin x = \dfrac{\sqrt{3}}{2}$, so $x = \dfrac{\pi}{3}$ or $x = \dfrac{2\pi}{3}$

$x = \dfrac{\pi}{3},\; \dfrac{2\pi}{3}$

Unit 8 — Polar & Complex Numbers

U8Polar Form & De Moivre's Theorem

🔑 Polar / Rectangular Conversion

$x = r\cos\theta,\; y = r\sin\theta,\; r = \sqrt{x^2+y^2},\; \theta = \arctan\!\left(\tfrac{y}{x}\right)$

De Moivre's Theorem:
$\bigl(r(\cos\theta + i\sin\theta)\bigr)^n = r^n(\cos n\theta + i\sin n\theta)$

Worked Example

Write $z = 1 + i$ in polar form.
$r = \sqrt{1^2+1^2} = \sqrt{2}$, $\theta = \arctan(1/1) = \pi/4$
$z = \sqrt{2}\!\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)$

$z = \sqrt{2}\,e^{i\pi/4}$

Instructions — Answer each question & submit individually

Problem 1

Angle Measure★☆☆☆☆

Convert $\dfrac{7\pi}{4}$ radians to degrees.

Problem 2

Unit Circle★☆☆☆☆

Find the exact value of $\sin\!\left(\dfrac{5\pi}{6}\right)$.

Problem 3

ASTC / Quadrant★★☆☆☆

If $\sin\theta = -\dfrac{3}{5}$ and $\theta$ is in Quadrant III, find the exact value of $\cos\theta$.

Problem 4

Arc Length★★☆☆☆

A circle has radius $r = 6$ cm. A central angle of $\dfrac{2\pi}{3}$ radians subtends an arc. Find the arc length $s$, in cm. Leave your answer in terms of $\pi$.

Problem 5

Trig Equations★★☆☆☆

Solve for $x$ in $[0, 2\pi)$:
$2\cos x + \sqrt{2} = 0$

Problem 6

Pythagorean Identity★★☆☆☆

Simplify: $\dfrac{\sin^2 x + \cos^2 x}{\sec^2 x - \tan^2 x}$

Problem 7

Graph — Amplitude/Period★★☆☆☆

State the amplitude, period, and phase shift of:
$y = -4\sin\!\left(3x - \dfrac{\pi}{2}\right) + 2$

Problem 8

Inverse Trig★★★☆☆

Evaluate exactly:
$\cos\!\left(\arcsin\!\left(\dfrac{5}{13}\right)\right)$

Problem 9

Double Angle★★★☆☆

Given $\sin\theta = \dfrac{3}{5}$ and $\theta \in \left(0, \dfrac{\pi}{2}\right)$, find the exact value of $\sin 2\theta$.

Problem 10

Sum Formula★★★☆☆

Use the angle addition formula to find the exact value of $\cos\!\left(75°\right)$. Express in simplified radical form.

Problem 11

Trig Equations★★★☆☆

Solve for $\theta \in [0, 2\pi)$:
$2\sin^2\theta - \sin\theta - 1 = 0$

Problem 12

Law of Cosines★★★☆☆

In triangle $ABC$, $a = 8$, $b = 6$, $c = 7$. Find $\cos A$ as an exact fraction.

Problem 13

Law of Sines★★★☆☆

In triangle $ABC$, $A = 30°$, $B = 45°$, $a = 10$. Find side $b$ (exact simplified radical form).

Problem 14

Identity Proof★★★★☆

Simplify the expression to a single trig function:
$\dfrac{\tan x + \cot x}{\sec x \csc x}$

Problem 15

Half Angle★★★★☆

Given $\cos\theta = \dfrac{7}{25}$ with $\theta \in (0, \pi/2)$, use the half-angle formula to find $\sin\!\left(\dfrac{\theta}{2}\right)$. Simplify completely.

Problem 16

Polar Coordinates★★★★☆

Convert the rectangular point $(-3, 3\sqrt{3})$ to polar form $(r, \theta)$ with $r > 0$ and $\theta \in [0, 2\pi)$.

Problem 17

Complex / De Moivre★★★★☆

Compute $\left(\sqrt{3} + i\right)^6$ using De Moivre's Theorem. Express in rectangular form $a + bi$.

Problem 18

Advanced Equation★★★★☆

Solve for $x \in [0, 2\pi)$:
$\cos 2x - \cos x = 0$

Problem 19

Triangle Area★★★★★

Triangle $ABC$ has sides $a = 5$, $b = 7$, $c = 8$. Using Heron's formula or otherwise, find the exact area of the triangle. Express in simplest radical form.

Problem 20

Sum-to-Product★★★★★

Prove the identity or find the value:
Simplify $\dfrac{\sin 3x - \sin x}{\cos 3x + \cos x}$ using sum-to-product formulas.
Express as a single trig function.

—

out of 20 questions

Complete Answer Key & Worked Solutions

1 — Angle Conversion315°

$\dfrac{7\pi}{4} \times \dfrac{180°}{\pi} = \dfrac{7 \times 180°}{4} = \dfrac{1260°}{4} = \mathbf{315°}$

2 — Unit Circle1/2

$\dfrac{5\pi}{6}$ is in Q2, reference angle $= \dfrac{\pi}{6}$. Since $\sin$ is positive in Q2: $\sin\!\left(\dfrac{5\pi}{6}\right) = \sin\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$

3 — Quadrant III Values−4/5

Use Pythagorean theorem on reference triangle: $\sin\theta = -3/5 \Rightarrow$ opp $= 3$, hyp $= 5$, adj $= 4$.
In Q3, $\cos$ is negative: $\cos\theta = -\dfrac{4}{5}$

4 — Arc Length4π cm

$s = r\theta = 6 \times \dfrac{2\pi}{3} = 4\pi$ cm

5 — Trig Equation3π/4, 5π/4

$2\cos x = -\sqrt{2} \Rightarrow \cos x = -\dfrac{\sqrt{2}}{2}$
Reference angle: $\dfrac{\pi}{4}$. $\cos$ is negative in Q2 and Q3.
$x = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$ and $x = \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}$

6 — Identity Simplification1

Numerator: $\sin^2 x + \cos^2 x = 1$ (Pythagorean identity)
Denominator: $\sec^2 x - \tan^2 x = 1$ (Pythagorean identity)
$\therefore \dfrac{1}{1} = \mathbf{1}$

7 — Graph ParametersA=4, P=2π/3, PS=π/6

$y = -4\sin\!\left(3x - \tfrac{\pi}{2}\right) + 2$: $A = |-4| = 4$,
Period $= \dfrac{2\pi}{3}$, Phase shift $= \dfrac{\pi/2}{3} = \dfrac{\pi}{6}$ (right), Vertical shift $= +2$

8 — Inverse Composition12/13

Let $\alpha = \arcsin(5/13)$, so $\sin\alpha = 5/13$ and $\alpha \in [-\pi/2, \pi/2]$ (Q1, positive).
By Pythagorean: $\cos\alpha = \dfrac{\sqrt{13^2 - 5^2}}{13} = \dfrac{\sqrt{144}}{13} = \dfrac{12}{13}$

9 — Double Angle Formula24/25

$\sin\theta = 3/5 \Rightarrow \cos\theta = 4/5$ (Q1, positive)
$\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \dfrac{3}{5} \cdot \dfrac{4}{5} = \dfrac{24}{25}$

10 — Angle Addition(√6−√2)/4

$75° = 45° + 30°$
$\cos 75° = \cos45°\cos30° - \sin45°\sin30°$
$= \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6}-\sqrt{2}}{4}$

11 — Quadratic Trigπ/6, 5π/6, 3π/2

Factor: $(2\sin\theta + 1)(\sin\theta - 1) = 0$
$\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}$ … wait, check: $(2(-1)+1)(-1-1)=(-1)(-2)=2\neq0$. Let's re-factor:
$2u^2 - u - 1 = (2u+1)(u-1)=0$ where $u=\sin\theta$
$\sin\theta = -\dfrac{1}{2} \Rightarrow \theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$ OR $\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}$
Solutions: $\theta = \dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

12 — Law of Cosines1/4

Law of Cosines: $a^2 = b^2 + c^2 - 2bc\cos A$
$64 = 36 + 49 - 2(6)(7)\cos A = 85 - 84\cos A$
$84\cos A = 85 - 64 = 21 \Rightarrow \cos A = \dfrac{21}{84} = \dfrac{1}{4}$

13 — Law of Sines10√2

$\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$: $\dfrac{b}{\sin 45°} = \dfrac{10}{\sin 30°}$
$b = 10 \cdot \dfrac{\sin 45°}{\sin 30°} = 10 \cdot \dfrac{\tfrac{\sqrt{2}}{2}}{\tfrac{1}{2}} = 10\sqrt{2}$

14 — Identity Simplification1

$\tan x + \cot x = \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \dfrac{1}{\sin x \cos x}$
$\sec x \csc x = \dfrac{1}{\cos x}\cdot\dfrac{1}{\sin x} = \dfrac{1}{\sin x \cos x}$
$\therefore \dfrac{1/(\sin x\cos x)}{1/(\sin x\cos x)} = \mathbf{1}$

15 — Half-Angle Formula3/5

$\sin\!\left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1-\cos\theta}{2}} = \sqrt{\dfrac{1 - 7/25}{2}} = \sqrt{\dfrac{18/25}{2}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}$
(Positive because $\theta/2 \in (0, \pi/4)$ is in Q1.)

16 — Polar Conversion(6, 2π/3)

$r = \sqrt{(-3)^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6$
Point is in Q2. $\tan\alpha = \dfrac{3\sqrt{3}}{3} = \sqrt{3}$, so ref angle $= \dfrac{\pi}{3}$.
$\theta = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$. Polar: $\left(6, \dfrac{2\pi}{3}\right)$

17 — De Moivre's Theorem−64

$\sqrt{3}+i$: $r = \sqrt{3+1} = 2$, $\theta = \arctan(1/\sqrt{3}) = \pi/6$
$(\sqrt{3}+i)^6 = 2^6\!\left(\cos\!\dfrac{6\pi}{6} + i\sin\!\dfrac{6\pi}{6}\right) = 64(\cos\pi + i\sin\pi)$
$= 64(-1 + 0i) = -64$

18 — Double-Angle Equation0, π/3, π, 5π/3

Substitute $\cos 2x = 2\cos^2 x - 1$:
$2\cos^2 x - 1 - \cos x = 0 \Rightarrow (2\cos x + 1)(\cos x - 1) = 0$
Wait, let's factor: $2u^2 - u - 1 = (2u+1)(u-1)=0$
$\cos x = 1 \Rightarrow x = 0$; $\cos x = -\dfrac{1}{2} \Rightarrow x = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$
Also re-check: $\cos 2x = \cos x \Rightarrow 2x = \pm x + 2k\pi$
$x = 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$ also check $2x = -x+2k\pi \Rightarrow 3x=2\pi k \Rightarrow x=0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$
Solutions: $x = 0,\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3}$

19 — Heron's Formula10√3

$s = \dfrac{5+7+8}{2} = 10$
$A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{10 \cdot 5 \cdot 3 \cdot 2} = \sqrt{300} = 10\sqrt{3}$

20 — Sum-to-Producttan 2x

Sum-to-product: $\sin A - \sin B = 2\cos\!\tfrac{A+B}{2}\sin\!\tfrac{A-B}{2}$, $\cos A + \cos B = 2\cos\!\tfrac{A+B}{2}\cos\!\tfrac{A-B}{2}$
Numerator: $\sin 3x - \sin x = 2\cos 2x \sin x$
Denominator: $\cos 3x + \cos x = 2\cos 2x \cos x$
$\dfrac{2\cos 2x \sin x}{2\cos 2x \cos x} = \dfrac{\sin x}{\cos x} = \tan x$
$= \tan x$

Pre-Calculus Trigonometry

Master Problem Set · All 8 Units · 20 Problems