# Trigonometry

March 13, 2020

## Trigonometry

### Relations

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

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

$$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$

$$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$

$$\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$$

$$\sin(a-b) = \sin(a)\cos(b) - \cos(a)\sin(b)$$

$$\tan(a+b) = \frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$$

$$\tan(a-b) = \frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$$

### Duplication

$$\cos(2a) = \cos^2(a)-\sin^2(a) = 1-2\sin^2(a) = 2\cos^2(a)-1$$

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

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

### from tangent

$$t = \tan(a/2)$$

$$\cos(a) = \frac{1-t^2}{1+t^2}$$

$$\sin(a) = \frac{2t}{1+t^2}$$

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

### Linearize

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

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

$$\cos(a)\cos(b) = \frac{1}{2}(\cos(a+b)+\cos(a-b))$$

$$\sin(a)\sin(b) = -\frac{1}{2}(\cos(a+b)-\cos(a-b))$$

$$\sin(a)\cos(b) = \frac{1}{2}(\sin(a+b)+\sin(a-b))$$

### Difference into product

$$\cos(p)+\cos(q) = 2\cos\frac{p+q}{2}\cos\frac{p-q}{2}$$

$$\cos(p)-\cos(q) = -2\sin\frac{p+q}{2}\sin\frac{p-q}{2}$$

$$\sin(p)+\sin(q) = 2\sin\frac{p+q}{2}\cos\frac{p-q}{2}$$

$$\sin(p)-\sin(q) = 2\cos\frac{p+q}{2}\sin\frac{p-q}{2}$$