## Related Formulae

$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)$
$\sin(2a)=2\sin(a)\cos(a)$
$\sin {\bigl (}{\tfrac {a}{2}}{\bigr )}=\pm {\sqrt {\frac {1-\cos(a)}{2}}}$

## Tangent Formulae

$\tan(a+b)={\frac {\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}}$
$\tan(a-b)={\frac {\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}}$
$\tan(2a)={\frac {2\tan(a)}{1-\tan ^{2}(a)}}={\frac {2\cot(a)}{\cot ^{2}(a)-1}}={\frac {2}{\cot(a)-\tan(a)}}$
$\tan {\bigl (}{\tfrac {a}{2}}{\bigr )}=\pm {\sqrt {\frac {1-\cos(a)}{1+\cos(a)}}}={\frac {\sin(a)}{1+\cos(a)}}={\frac {1-\cos(a)}{\sin(a)}}={\frac {-1\pm {\sqrt {1+\tan ^{2}(a)}}}{\tan(a)}}$

In the last row of expressions, if $0^{\circ }\leq a\leq 90^{\circ }$  then the trigonometric functions are all positive so the positive sign is needed before the square root.

## Derivations

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

Using cofunctions we know that $\sin(a)=\cos(90^{\circ }-a)$  . Use the formula for $\cos(a-b)$  and cofunctions we can write

 $\sin(a+b)$ $=\cos(90-(a+b))$ $=\cos {\bigl (}(90^{\circ }-a)-b{\bigr )}$ $=\cos(90^{\circ }-a)\cos(b)+\sin(90^{\circ }-a)\sin(b)$ $={\color {red}\sin(a)\cos(b)+\cos(a)\sin(b)}$ • $\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)$

Having derived $\sin(a+b)$  we replace $b$  with $-b$  and use the fact that cosine is even and sine is odd.

 $\sin {\bigl (}a+(-b){\bigr )}$ $=\sin(a)\cos(-b)+\cos(a)\sin(-b)$ $=\sin(a)\cos(b)+\cos(a){\bigl (}-\sin(b){\bigr )}$ $={\color {red}\sin(a)\cos(b)-\cos(a)\sin(b)}$ ## Related Formulae

$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$
$\cos(2a)=\cos ^{2}(a)-\sin ^{2}(a)=2\cos ^{2}(a)-1=1-2\sin ^{2}(a)$
$\cos {\bigl (}{\tfrac {a}{2}}{\bigr )}=\pm {\sqrt {\frac {1+\cos(a)}{2}}}$

## Derivations

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

Using $\cos(a+b)$  and the fact that cosine is even and sine is odd, we have

 $\cos {\bigl (}a+(-b){\bigr )}$ $=\cos(a)\cos(-b)-\sin(a)\sin(-b)$ $=\cos(a)\cos(b)-\sin(a){\bigl (}-\sin(b){\bigr )}$ $={\color {red}\cos(a)\cos(b)+\sin(a)\sin(b)}$ 