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Trigonometry

March 13, 2020

Trigonometry

Relations

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

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

Addition

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

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

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

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

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

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

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

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

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

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

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