# Trigonometry/For Enthusiasts/Less-Used Trig Identities

## Triangle Identities

In addition to the Law of Sines, the Law of Cosines, and the Law of Tangents, there are numerous other identities that apply to the three angles A, B, and C of any triangle (where A+B+C=180° and each of A, B, and C is greater than zero). Some of the most notable ones follow:

${\displaystyle \cos ^{2}(A)+\cos ^{2}(B)+\cos ^{2}(C)+2\cos(A)\cos(B)\cos(C)=1}$
${\displaystyle \sin(A)+\sin(B)+\sin(C)=4\cos {\bigl (}{\tfrac {A}{2}}{\bigr )}\cos {\bigl (}{\tfrac {B}{2}}{\bigr )}\cos {\bigl (}{\tfrac {C}{2}}{\bigr )}}$
${\displaystyle \tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)}$
${\displaystyle \tan {\bigl (}{\tfrac {A}{2}}{\bigr )}\tan {\bigl (}{\tfrac {B}{2}}{\bigr )}+\tan {\bigl (}{\tfrac {B}{2}}{\bigr )}\tan {\bigl (}{\tfrac {C}{2}}{\bigr )}+\tan {\bigl (}{\tfrac {C}{2}}{\bigr )}\tan {\bigl (}{\tfrac {A}{2}}{\bigr )}=1}$
${\displaystyle \cot(A)\cot(B)+\cot(B)\cot(C)+\cot(C)\cot(A)=1}$
${\displaystyle \cot {\bigl (}{\tfrac {A}{2}}{\bigr )}\cot {\bigl (}{\tfrac {B}{2}}{\bigr )}\cot {\bigl (}{\tfrac {C}{2}}{\bigr )}=\cot {\bigl (}{\tfrac {A}{2}}{\bigr )}+\cot {\bigl (}{\tfrac {B}{2}}{\bigr )}+\cot {\bigl (}{\tfrac {C}{2}}{\bigr )}}$
${\displaystyle \sin(A)\sin(B)\sin(C)={\frac {1}{{\bigl (}\cot(A)+\cot(B){\bigr )}{\bigl (}\cot(B)+\cot(C){\bigr )}{\bigl (}\cot(C)+\cot(A){\bigr )}}}}$
${\displaystyle {\frac {\sin(A)+\sin(B)-\sin(C)}{\sin(A)+\sin(B)+\sin(C)}}=\tan {\bigl (}{\tfrac {A}{2}}{\bigr )}\tan {\bigl (}{\tfrac {B}{2}}{\bigr )}}$

## Pythagoras

${\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1}$
${\displaystyle 1+\tan ^{2}(x)=\sec ^{2}(x)}$
${\displaystyle 1+\cot ^{2}(x)=\csc ^{2}(x)}$

These are all direct consequences of Pythagoras's theorem.

## Sum/Difference of angles

${\displaystyle \cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)}$
${\displaystyle \sin(x\pm y)=\sin(x)\cos(y)\pm \sin(y)\cos(x)}$
${\displaystyle \tan(x\pm y)={\frac {\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}}}$

## Product to Sum

${\displaystyle 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y)}$
${\displaystyle 2\cos(x)\cos(y)=\cos(x-y)+\cos(x+y)}$
${\displaystyle 2\sin(x)\cos(y)=\sin(x-y)+\sin(x+y)}$

## Sum and difference to product

${\displaystyle A\sin(x)+B\cos(x)=C\sin(x+y)}$ , where ${\displaystyle C={\sqrt {A^{2}+B^{2}}}}$ and ${\displaystyle y=\pm \arctan {\bigl (}{\tfrac {B}{A}}{\bigr )}}$
${\displaystyle \sin(\alpha )+\sin(\beta )=2\sin {\bigl (}{\tfrac {\alpha +\beta }{2}}{\bigr )}\cos {\bigl (}{\tfrac {\alpha -\beta }{2}}{\bigr )}}$
${\displaystyle \sin(\alpha )-\sin(\beta )=2\cos {\bigl (}{\tfrac {\alpha +\beta }{2}}{\bigr )}\sin {\bigl (}{\tfrac {\alpha -\beta }{2}}{\bigr )}}$
${\displaystyle \cos(\alpha )+\cos(\beta )=2\cos {\bigl (}{\tfrac {\alpha +\beta }{2}}{\bigr )}\cos {\bigl (}{\tfrac {\alpha -\beta }{2}}{\bigr )}}$
${\displaystyle \cos(\alpha )-\cos(\beta )=-2\sin {\bigl (}{\tfrac {\alpha +\beta }{2}}{\bigr )}\sin {\bigl (}{\tfrac {\alpha -\beta }{2}}{\bigr )}}$

## Multiple angle

${\displaystyle \cos(2x)=\cos ^{2}(x)-\sin ^{2}(x)=2\cos ^{2}(x)-1=1-2\sin ^{2}(x)}$
${\displaystyle \sin(2x)=2\sin(x)\cos(x)}$
${\displaystyle \tan(2x)={\frac {2\tan(x)}{1-\tan ^{2}(x)}}}$
${\displaystyle \cot(2x)={\frac {\cot(x)-\tan(x)}{2}}}$
${\displaystyle \csc(2x)={\frac {\cot(x)+\tan(x)}{2}}}$
${\displaystyle \cos(3x)=4\cos ^{3}(x)-3\cos(x)}$
${\displaystyle \sin(3x)=-4\sin ^{3}(x)+3\sin(x)}$
${\displaystyle \tan(3x)={\frac {3\tan(x)-\tan ^{3}(x)}{1-3\tan ^{2}(x)}}}$
${\displaystyle \cos(4x)=8\cos ^{4}(x)-8\cos ^{2}(x)+1}$
${\displaystyle \sin(4x)=4\sin(x)\cos ^{3}(x)-4\sin ^{3}(x)\cos(x)}$
${\displaystyle \sin ^{2}(4x)=16{\Big [}\sin ^{2}(x)-5\sin ^{4}(x)+8\sin ^{6}(x)-4\sin ^{8}(x){\Big ]}}$
${\displaystyle \tan(4x)={\frac {4\tan(x)-4\tan ^{3}(x)}{1-6\tan ^{2}(x)+\tan ^{4}(x)}}}$
${\displaystyle \cos(5x)=16\cos ^{5}(x)-20\cos ^{3}(x)+5\cos(x)}$
${\displaystyle \sin(5x)=16\sin ^{5}(x)-20\sin ^{3}(x)+5\sin(x)}$
${\displaystyle \tan(5x)={\frac {5\tan(x)-10\tan ^{3}(x)+\tan ^{5}(x)}{1-10\tan ^{2}(x)+5\tan ^{4}(x)}}}$
${\displaystyle \cos(6x)=32\cos ^{6}(x)-48\cos ^{4}(x)+18\cos ^{2}(x)-1}$
${\displaystyle \cos(7x)=64\cos ^{7}(x)-112\cos ^{5}(x)+56\cos ^{3}(x)-7\cos(x)}$
${\displaystyle \sin(7x)=-64\sin ^{7}(x)+112\sin ^{5}(x)-56\sin ^{3}(x)+7\sin(x)}$
${\displaystyle \cos(8x)=128\cos ^{8}(x)-256\cos ^{6}(x)+160\cos ^{4}(x)-32\cos ^{2}(x)+1}$
${\displaystyle \cos(nx)=2\cos(x)\cos {\bigl (}(n-1)x{\bigr )}-\cos {\bigl (}(n-2)x{\bigr )}}$
${\displaystyle \sin(nx)=2\cos(x)\sin {\bigl (}(n-1)x{\bigr )}-\sin {\bigl (}(n-2)x{\bigr )}}$

These are all direct consequences of the sum/difference formulae

## Half angle

${\displaystyle \cos {\bigl (}{\tfrac {x}{2}}{\bigr )}=\pm {\sqrt {\frac {1+\cos(x)}{2}}}}$
${\displaystyle \sin {\bigl (}{\tfrac {x}{2}}{\bigr )}=\pm {\sqrt {\frac {1-\cos(x)}{2}}}}$
${\displaystyle \tan {\bigl (}{\tfrac {x}{2}}{\bigr )}={\frac {1-\cos(x)}{\sin(x)}}={\frac {\sin(x)}{1+\cos(x)}}=\pm {\sqrt {\frac {1-\cos(x)}{1+\cos(x)}}}}$
${\displaystyle \cos ^{2}{\bigl (}{\tfrac {3x}{2}}{\bigr )}=2\cos ^{3}(x)-{\frac {3\cos(x)+1}{2}}}$

In cases with ${\displaystyle \pm }$ , the sign of the result must be determined from the value of ${\displaystyle {\frac {x}{2}}}$ . These derive from the ${\displaystyle \cos(2x)}$ formulae.

## Power Reduction

${\displaystyle \sin ^{2}(\theta )={\frac {1-\cos 2\theta }{2}}}$
${\displaystyle \cos ^{2}(\theta )={\frac {1+\cos 2\theta }{2}}}$
${\displaystyle \tan ^{2}(\theta )={\frac {1-\cos 2\theta }{1+\cos 2\theta }}}$

## Even/Odd

${\displaystyle \sin(-\theta )=-\sin(\theta )}$
${\displaystyle \cos(-\theta )=\cos(\theta )}$
${\displaystyle \tan(-\theta )=-\tan(\theta )}$
${\displaystyle \csc(-\theta )=-\csc(\theta )}$
${\displaystyle \sec(-\theta )=\sec(\theta )}$
${\displaystyle \cot(-\theta )=-\cot(\theta )}$

## Calculus

${\displaystyle {\frac {d}{dx}}[\sin(x)]=\cos(x)}$
${\displaystyle {\frac {d}{dx}}[\cos(x)]=-\sin(x)}$
${\displaystyle {\frac {d}{dx}}[\tan(x)]=\sec ^{2}(x)}$
${\displaystyle {\frac {d}{dx}}[\sec(x)]=\sec(x)\tan(x)}$
${\displaystyle {\frac {d}{dx}}[\csc(x)]=-\csc(x)\cot(x)}$
${\displaystyle {\frac {d}{dx}}[\cot(x)]=-\csc ^{2}(x)}$