how to evaluate trig functions with sum and difference identities

the sum and difference trig identities are useful in evaluating more stuff

the proof for these are quite intense, better not to worry

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

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

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

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

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

worked example: evaluate $\sin (\frac{\pi }{4}+\frac{\pi }{3})$
$\sin (a)\cdot \cos (b)+\cos (a)\cdot \sin (b)$
$\sin (\frac{\pi }{4})\cdot \cos (\frac{\pi }{3})+\cos (\frac{\pi }{4})\cdot \sin (\frac{\pi }{3})$
$\frac{\sqrt{2}}{2}\cdot \frac{1}{2}+\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}$
$\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}$
$\frac{\sqrt{2}+\sqrt{6}}{4}$

worked example: evaluate $\cos (\frac{5\pi }{12})$
$\cos (\frac{\pi }{4}+\frac{\pi }{6})$
$\cos (a)\cdot \cos (b)-\sin (a)\cdot \sin (b)$
$\cos (\frac{\pi }{4})\cdot \cos (\frac{\pi }{6})-\sin (\frac{\pi }{4})\cdot \sin (\frac{\pi }{6})$
$\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot \frac{1}{2}$
$\frac{\sqrt{6}-\sqrt{2}}{4}$

worked example: evaluate $\sin (-\frac{\pi }{12})$
$\sin (\frac{\pi }{4}-\frac{\pi }{3})$
$\sin (a)\cdot \cos (b)-\cos (a)\cdot \sin (b)$
$\sin (\frac{\pi }{4})\cdot \cos (\frac{\pi }{3})-\cos (\frac{\pi }{4})\cdot \sin (\frac{\pi }{3})$
$\frac{\sqrt{2}}{2}\cdot \frac{1}{2}-\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}$
$\frac{\sqrt{2}-\sqrt{6}}{4}$

worked example: evaluate $\cos (\frac{7\pi }{12})$
$\cos (\frac{\pi }{3}+\frac{\pi }{4})$
$\cos (a)\cdot \cos (b)-\sin (a)\cdot \sin (b)$
$\cos (\frac{\pi }{3})\cdot \cos (\frac{\pi }{4})-\sin (\frac{\pi }{3})\cdot \sin (\frac{\pi }{4})$
$\frac{1}{2}\cdot \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}$
$\frac{\sqrt{2}-\sqrt{6}}{4}$

worked example: evaluate $\tan (\frac{\pi }{12})$
$\tan (\frac{\pi }{3}-\frac{\pi }{4})$
$\frac{\tan (a)-\tan (b)}{1+\tan (a)\cdot \tan (b)}$
$\frac{\sqrt{3}-1}{1+\sqrt{3}}$
rationalize
$\frac{\sqrt{3}-1}{1+\sqrt{3}}\cdot \frac{1-\sqrt{3}}{1-\sqrt{3}}$
$\frac{(\sqrt{3}-1)(1-\sqrt{3})}{1-3}$
$\frac{\sqrt{3}-3-1+\sqrt{3}}{1-3}$
$\frac{2\sqrt{3}-4}{-2}$
$2-\sqrt{3}$

worked example: evaluate $\tan (-\frac{7\pi }{12})$
$\tan (-\frac{\pi }{3}+-\frac{\pi }{4})$
$\frac{\tan (a)+\tan (b)}{1-\tan (a)\cdot \tan (b)}$
$\tan (-\frac{\pi }{3})=-\sqrt{3}$
$\tan (-\frac{\pi }{4})=-1$
$\frac{-\sqrt{3}-1}{1-\sqrt{3}}$
rationalize
$\frac{-\sqrt{3}-1}{1-\sqrt{3}}\cdot \frac{1+\sqrt{3}}{1+\sqrt{3}}$
$\frac{(-\sqrt{3}-1)(1+\sqrt{3})}{1-3}$
$\frac{-\sqrt{3}-3-1-\sqrt{3}}{1-3}$
$\frac{-2\sqrt{3}-4}{-2}$
$\sqrt{3}+2$

worked example: given angles $a$ and $b$ in quadrant 1, $\sin (a)=0.7$ and $\cos (b)=0.8$, find $\cos (a-b)$
$\cos (a)\cdot \cos (b)+\sin (a)\cdot \sin (b)$
make a triangle to use $\sin (a)=0.7$ to find $\cos (a)$

$\cos (a)=\frac{\sqrt{51}}{10}$
make a triangle to use $\cos (b)=0.8$ to find $\sin (b)$

$\sin (b)=\frac{6}{10}$
$\frac{\sqrt{51}}{10}\cdot 0.8+0.7\ast \frac{6}{10}$
$\frac{8\sqrt{51}+42}{100}$
$\frac{4\sqrt{51}+21}{50}$

worked example: given angles $a$ and $b$ in quadrant 1, $\sin (a)=0.7$ and $\cos (b)=0.8$, find $\tan (a+b)$
$\frac{\tan (a)+\tan (b)}{1-\tan (a)\cdot \tan (b)}$
make a triangle to use $\sin (a)=0.7$ to find $\tan (a)$
make a triangle to use $\cos (b)=0.8$ to find $\tan (b)$

$\tan (a)=\frac{7}{\sqrt{51}}$
$\tan (b)=\frac{3}{4}$
$\frac{\frac{7}{\sqrt{51}}+\frac{3}{4}}{1-\frac{7}{\sqrt{51}}\cdot \frac{3}{4}}$
i have no idea how to simplify this further :/