Precalculus Trigonometry

01

Angle Measurement & the Unit Circle

Degree–Radian Conversion

\(\text{radians} = \text{degrees} \times \dfrac{\pi}{180}\) \(\text{degrees} = \text{radians} \times \dfrac{180}{\pi}\)

★ Memorize — Key Angles

\(0°=0,\ 30°=\tfrac{\pi}{6},\ 45°=\tfrac{\pi}{4},\ 60°=\tfrac{\pi}{3},\ 90°=\tfrac{\pi}{2},\ 180°=\pi,\ 270°=\tfrac{3\pi}{2},\ 360°=2\pi\)

Example

Convert \(150°\) to radians.

\(150 \times \dfrac{\pi}{180} = \dfrac{5\pi}{6}\)

Answer: \(\dfrac{5\pi}{6}\)

02

Trig Functions & Unit Circle Values

Definitions on Unit Circle \((x,y)\)

\(\sin\theta = y,\quad \cos\theta = x,\quad \tan\theta = \dfrac{y}{x}\ (x\neq0)\)

★ Memorize — Exact Values

\(\theta\)	\(\sin\theta\)	\(\cos\theta\)	\(\tan\theta\)
\(0\)	\(0\)	\(1\)	\(0\)
\(\frac{\pi}{6}\)	\(\frac{1}{2}\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{\sqrt{3}}\)
\(\frac{\pi}{4}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{\sqrt{2}}{2}\)	\(1\)
\(\frac{\pi}{3}\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{2}\)	\(\sqrt{3}\)
\(\frac{\pi}{2}\)	\(1\)	\(0\)	undef

★ Reference Angle Signs by Quadrant

Q1: all +, Q2: sin+, Q3: tan+, Q4: cos+ (mnemonic: All Students Take Calculus)

Example

Find \(\sin\!\left(\dfrac{5\pi}{6}\right)\).

Reference angle: \(\pi - \dfrac{5\pi}{6} = \dfrac{\pi}{6}\). Quadrant II → sin is positive. \(\sin\!\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

Answer: \(\dfrac{1}{2}\)

03

Pythagorean & Reciprocal Identities

Core Identities — MUST Memorize

\(\sin^2\theta + \cos^2\theta = 1\)
\(1 + \tan^2\theta = \sec^2\theta\)
\(1 + \cot^2\theta = \csc^2\theta\)

Derived Forms

\(\sin^2\theta = 1 - \cos^2\theta \qquad \cos^2\theta = 1 - \sin^2\theta\)

Example

Given \(\tan x = 2\), find \(\sin^2 x\).

Use \(1+\tan^2 x = \sec^2 x \Rightarrow \sec^2 x = 5 \Rightarrow \cos^2 x = \tfrac{1}{5}\).
Then \(\sin^2 x = 1 - \tfrac{1}{5} = \tfrac{4}{5}\).

Answer: \(\dfrac{4}{5}\)

04

Graphing Trig Functions

Standard Form \(y = A\sin(Bx - C) + D\)

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

★ Range

Range of \(y = A\sin(\ldots)+D\) is \([D-|A|,\ D+|A|]\)

Example

For \(y = 2\sin(3x - \tfrac{\pi}{4}) + 1\), find the period and phase shift.

Period \(=\dfrac{2\pi}{3}\), Phase shift \(=\dfrac{\pi/4}{3}=\dfrac{\pi}{12}\) (right)

Period: \(\dfrac{2\pi}{3}\), Phase shift: \(\dfrac{\pi}{12}\)

05

Double-Angle & Half-Angle Formulas

Double-Angle

\(\sin 2\theta = 2\sin\theta\cos\theta\)
\(\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1\)

Half-Angle

\(\sin^2\theta = \dfrac{1-\cos 2\theta}{2} \qquad \cos^2\theta = \dfrac{1+\cos 2\theta}{2}\)

06

Sum & Difference Formulas

Sum & Difference

\(\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\)
\(\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta\)

Example

Given \(\sin\alpha=\tfrac{1}{2}\), \(\cos\beta=\tfrac{\sqrt{3}}{2}\) (both in Q1), find \(\sin(\alpha+\beta)\).

\(\cos\alpha=\tfrac{\sqrt{3}}{2},\ \sin\beta=\tfrac{1}{2}\)
\(\sin(\alpha+\beta)=\tfrac{1}{2}\cdot\tfrac{\sqrt{3}}{2}+\tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2}=\tfrac{\sqrt{3}}{4}+\tfrac{\sqrt{3}}{4}=\tfrac{\sqrt{3}}{2}\)

Answer: \(\dfrac{\sqrt{3}}{2}\)

07

Law of Sines & Law of Cosines

Law of Sines

\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)

Law of Cosines

\(a^2 = b^2 + c^2 - 2bc\cos A\)

Triangle Area

\(\text{Area} = \dfrac{1}{2}bc\sin A\)

08

Inverse Trigonometric Functions

Domains & Ranges

\(\arcsin: [-1,1]\to[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\)
\(\arccos: [-1,1]\to[0,\pi]\)
\(\arctan: (-\infty,\infty)\to(-\tfrac{\pi}{2},\tfrac{\pi}{2})\)

★ Key Values

\(\arcsin(\tfrac{1}{2})=\tfrac{\pi}{6},\quad \arccos(0)=\tfrac{\pi}{2},\quad \arctan(1)=\tfrac{\pi}{4}\)

20 Exam-Style Problems