Notes can be found as interactive webpage at

Trig Identities

Reciprocal #

$$\sin(x)=\frac{1}{\csc(x)}$$

$$\cos(x)=\frac{1}{\sec(x)}$$

$$\tan(x)=\frac{\sin(x)}{\cos(x)}=\frac{1}{\cot(x)}$$

Pythagorean #

$$\sin^2(x) + \cos^2(x) = 1$$

$$1+\tan^2(x) = \sec^2(x)$$

$$1+\cot^2(x)=\csc^2(x)$$

Cofunction #

$$\sin\Big(\frac{\pi}{2}-x\Big) = \cos(x)$$ $$\csc\Big(\frac{\pi}{2}-x\Big) = \sec(x)$$

$$\cos\Big(\frac{\pi}{2}-x\Big) = \sin(x)$$ $$\sec\Big(\frac{\pi}{2}-x\Big) = \csc(x)$$

$$\tan\Big(\frac{\pi}{2}-x\Big) = \cot(x)$$ $$\cot\Big(\frac{\pi}{2}-x\Big) = \tan(x)$$

Even/Odd #

$$ \sin(-x) = -\sin(x)$$ $$ \csc(-x) = -\csc(x)$$

$$ \cos(-x) = \cos(x)$$ $$ \sec(-x) = \sec(x)$$

$$ \tan(-x) = - \tan(x)$$ $$ \cot(-x) = -\cot(x)$$

Bonus fact: .$\int_{-A}^A \text{[odd]}(x)\ dx = 0$; .$\int_{-A}^A \text{[even]}(x)\ dx = \int_0^A \text{[even]}(x)\ dx$

Sum and Difference #

$$ \sin(u \pm v) = \sin(u) \cdot \cos(v) \pm \cos(u) \cdot \sin(v)$$ $$ \cos(u \pm v) = \cos(u) \cdot \cos(v) \pm \sin(u) \cdot \sin(v)$$ $$ \tan(u \pm v) = \frac{\tan(u) \pm \tan(v)}{1 \mp \tan(u) \tan(v)}$$

Double-Angle #

$$ \sin(2u) = 2 \sin(u) \cos(u)$$

$$ \cos(2u) = 2 \cos^2(u) - 1$$ $$ … = 1- 2 \sin^2(u) $$ $$ … = \cos^2(u) - \sin^2(u) $$

$$ \tan(2u) = \frac{2 \tan(u)}{1 - \tan^2(u)}$$

Power Reducing #

$$\sin^2(u) = \frac{1 - \cos(2u)}{2}$$

$$\cos^2(u) = \frac{1 + \cos(2u)}{2}$$

$$\tan^2(u) = \frac{1 - \cos(2u)}{1 + \cos(2u)}$$

Sum-to-Product #

$$\begin{align*} \sin(u) + \sin(v) &= 2\sin\bigg(\frac{u + v}{2}\bigg) \cos\bigg(\frac{u - v}{2}\bigg) \\ \sin(u) - \sin(v) &= 2\cos\bigg(\frac{u + v}{2}\bigg) \sin\bigg(\frac{u - v}{2}\bigg) \\ \cos(u) + \cos(v) &= 2\cos\bigg(\frac{u + v}{2}\bigg) \cos\bigg(\frac{u - v}{2}\bigg) \\ \cos(u) - \cos(v) &= -2\sin\bigg(\frac{u + v}{2}\bigg) \sin\bigg(\frac{u - v}{2}\bigg) \end{align*}$$

Product-to-Sum #

$$\begin{align*} \sin(u) \sin(v) &= \frac{1}{2}\Big[\cos(u - v) - \cos(u + v)\Big] \\ \sin(u) \cos(v) &= \frac{1}{2}\Big[\sin(u + v) + \sin(u - v)\Big] \\ \cos(u) \sin(v) &= \frac{1}{2}\Big[\sin(u + v) - \sin(u - v)\Big] \\ \cos(u) \cos(v) &= \frac{1}{2}\Big[\cos(u - v) + \cos(u + v)\Big] \\ \end{align*}$$