## Related Formulae

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

## Tangent Formulae

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

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

## Derivations

• ${\displaystyle \sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)}$

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

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

• ${\displaystyle \sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)}$

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

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

## Related Formulae

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

## Derivations

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

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

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