An angle is formed by rotating a ray from an initial side to a terminal side.
• Degree: Full circle = 360°
• Radian: Full circle = 2π rad
• Conversion: rad = deg × π/180
\(\theta(\text{rad}) = \theta(°)\times\dfrac{\pi}{180}\)
\(\theta(°) = \theta(\text{rad})\times\dfrac{180}{\pi}\)
②
Unit Circle & Basic Trig Functions
For a point \((x,y)\) on the unit circle (radius 1) at angle \(\theta\):
\(\sin\theta = y\), \(\cos\theta = x\), \(\tan\theta = \dfrac{y}{x}\)
\(\csc\theta=\frac{1}{\sin\theta}\quad \sec\theta=\frac{1}{\cos\theta}\quad \cot\theta=\frac{1}{\tan\theta}\)
Three fundamental identities derived from \(x^2+y^2=1\):
\(\sin^2\theta+\cos^2\theta=1\)
\(1+\tan^2\theta=\sec^2\theta\)
\(1+\cot^2\theta=\csc^2\theta\)
④
Angle Addition & Double Angle
\(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\)
\(\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\)
\(\sin 2A=2\sin A\cos A\)
\(\cos 2A=\cos^2\!A-\sin^2\!A=2\cos^2\!A-1=1-2\sin^2\!A\)
\(\tan 2A=\dfrac{2\tan A}{1-\tan^2 A}\)
⑤
Graphs of Trig Functions
General form: \(y=A\sin(Bx+C)+D\)
• Amplitude = \(|A|\)
• Period = \(\dfrac{2\pi}{|B|}\) (for sin/cos), \(\dfrac{\pi}{|B|}\) (for tan)
• Phase shift = \(-\dfrac{C}{B}\)
• Vertical shift = \(D\)
⑥
Inverse Trig Functions & Laws
\(\sin^{-1}x\): domain \([-1,1]\), range \(\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\)
\(\cos^{-1}x\): domain \([-1,1]\), range \([0,\pi]\)
\(\tan^{-1}x\): domain \(\mathbb{R}\), range \(\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\)
\textbf{Law of Sines:}\; \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)
\(\textbf{Law of Cosines:}\; c^2=a^2+b^2-2ab\cos C\)
Example 2 — Pythagorean Identity
If \(\sin\theta=\dfrac{3}{5}\) and \(\theta\) is in Quadrant I, find \(\cos\theta\).
\(\cos^2\theta = 1-\sin^2\theta = 1-\dfrac{9}{25}=\dfrac{16}{25}\)
Since QI: \(\cos\theta=\dfrac{4}{5}\)
Answer: cos θ = 4/5
Example 3 — Double Angle Formula
If \(\sin\theta=\dfrac{3}{5}\) (QI), find \(\sin 2\theta\).
\(\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5}=\dfrac{24}{25}\)
Answer: 24/25
Example 4 — Period of a Function
Find the period of \(y=3\sin(2x-\pi)+1\).
\(B=2\Rightarrow \text{Period}=\dfrac{2\pi}{|B|}=\dfrac{2\pi}{2}=\pi\)
Answer: π
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