Lijst van goniometrische gelijkheden

${\displaystyle \cos ^{2}(x)+\sin ^{2}(x)=1}$

${\displaystyle {\begin{array}{rl}\tan x=&{\cfrac {\sin x}{\cos x}}\\\sec x=&{\cfrac {1}{\cos x}}\\\csc x=&{\cfrac {1}{\sin x}}\\\cot x=&{\cfrac {1}{\tan x}}={\cfrac {\cos x}{\sin x}}\\\end{array}}}$

${\displaystyle {\begin{array}{rlrl}\sin x=&\sin(x+2\pi )=\sin(\pi -x)&\sin x=&\cos \left({\frac {\pi }{2}}-x\right)\\\cos x=&\cos(x+2\pi )=\cos(-x)&\cos x=&\sin \left({\frac {\pi }{2}}-x\right)\\\tan x=&\tan(x+2\pi )=\tan(x+\pi )&\tan x=&\cot \left({\frac {\pi }{2}}-x\right)\\-\sin x=&\sin(x+\pi )=\sin(-x)&\\-\cos x=&\cos(x+\pi )=\cos(\pi -x)&\\-\tan x=&\tan(-x)=\tan(\pi -x)&\end{array}}}$

De volgende drie identiteiten zijn gebaseerd op de stelling van Pythagoras. De tweede en derde zijn af te leiden uit de bovenste formule door te delen door het kwadraat van de cosinus en respectievelijk het kwadraat van de sinus.

${\displaystyle {\begin{array}{ccccc}\cos ^{2}x&+&\sin ^{2}x&=&1\\1&+&\tan ^{2}x&=&\sec ^{2}x&=&{\cfrac {1}{\cos ^{2}x}}\\\cot ^{2}x&+&1&=&\csc ^{2}x&=&{\cfrac {1}{\sin ^{2}x}}\\\end{array}}}$

${\displaystyle {\begin{array}{rcl}\sin(x\pm y)&=&\sin(x)\cos(y)\pm \cos(x)\sin(y)\\\cos(x\pm y)&=&\cos(x)\cos(y)\mp \sin(x)\sin(y)\\\tan(x\pm y)&=&{\cfrac {\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}}\\\end{array}}}$

${\displaystyle {\begin{array}{rcl}\sin(2x)&=&2\sin(x)\cos(x)={\cfrac {2\tan(x)}{1+\tan ^{2}(x)}}\\\cos(2x)&=&\cos ^{2}(x)-\sin ^{2}(x)=2\cos ^{2}(x)-1=1-2\sin ^{2}(x)={\cfrac {1-\tan ^{2}(x)}{1+\tan ^{2}(x)}}\\\tan(2x)&=&{\cfrac {2\tan(x)}{1-\tan ^{2}(x)}}\\\end{array}}}$

${\displaystyle {\begin{array}{rcl}\sin(3x)&=&3\sin(x)-4\sin ^{3}(x)\\\cos(3x)&=&4\cos ^{3}(x)-3\cos(x)\\\tan(3x)&=&{\cfrac {3\tan(x)-\tan ^{3}(x)}{1-3\tan ^{2}(x)}}\end{array}}}$

${\displaystyle {\begin{array}{ccl}\cos ^{2}(x)&=&{\tfrac {1}{2}}(1+\cos(2x))\\\sin ^{2}(x)&=&{\tfrac {1}{2}}(1-\cos(2x))\\\sin ^{2}(x)\cos ^{2}(x)&=&{\tfrac {1}{8}}(1-\cos(4x))\end{array}}}$

Met de t-formules, zo genoemd vanwege de substitutie:

${\displaystyle t=\tan({\tfrac {1}{2}}x)}$

zijn vergelijkingen met goniometrische identiteiten in ${\displaystyle x}$ op te lossen door ze eerst te schrijven als functie van ${\displaystyle t}$ en later weer terug te transformeren naar ${\displaystyle x}$. Er geldt:

${\displaystyle \tan(x)={\frac {2t}{1-t^{2}}}}$

${\displaystyle \cos(x)={\frac {1-t^{2}}{1+t^{2}}}}$

${\displaystyle \sin(x)={\frac {2t}{1+t^{2}}}}$

${\displaystyle {\begin{array}{rcl}\cos \left({\tfrac {1}{2}}x\right)&=&\pm \,{\sqrt {\cfrac {1+\cos x}{2}}}\\\sin \left({\tfrac {1}{2}}x\right)&=&\pm \,{\sqrt {\cfrac {1-\cos x}{2}}}\\\tan \left({\tfrac {1}{2}}x\right)&=&{\cfrac {\sin({\tfrac {1}{2}}x)}{\cos({\tfrac {1}{2}}x)}}=\pm \,{\sqrt {\cfrac {1-\cos x}{1+\cos x}}}\\\tan \left({\tfrac {1}{2}}x\right)&=&{\cfrac {\sin x}{1+\cos x}}={\cfrac {1-\cos x}{\sin x}}={\cfrac {-1\pm {\sqrt {1+\tan ^{2}x}}}{\tan x}}\\\end{array}}}$

${\displaystyle {\begin{array}{ccrrr}\sin(x)+\sin(y)&=&2\sin {\cfrac {x+y}{2}}\cos {\cfrac {x-y}{2}}\\\sin(x)-\sin(y)&=&2\cos {\cfrac {x+y}{2}}\sin {\cfrac {x-y}{2}}\\\cos(x)+\cos(y)&=&2\cos {\cfrac {x+y}{2}}\cos {\cfrac {x-y}{2}}\\\cos(x)-\cos(y)&=&-2\sin {\cfrac {x+y}{2}}\sin {\cfrac {x-y}{2}}\\\end{array}}}$

${\displaystyle {\begin{array}{ccc}\cos(x)\cos(y)&=&{\cfrac {\cos(x-y)+\cos(x+y)}{2}}\\\sin(x)\sin(y)&=&{\cfrac {\cos(x-y)-\cos(x+y)}{2}}\\\sin(x)\cos(y)&=&{\cfrac {\sin(x-y)+\sin(x+y)}{2}}\\\end{array}}}$