Lijst van goniometrische gelijkheden

De goniometrische basisfuncties zijn op diverse manieren met elkaar verbonden. Dit artikel geeft een lijst met goniometrische gelijkheden.

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

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

Dit is de grondformule van de goniometrie en is gebaseerd op de stelling van Pythagoras. De tweede en derde zijn hieruit af te leiden door te delen door het kwadraat van de cosinus en sinus.

${\displaystyle \tan ^{2}x+1={\frac {1}{\cos ^{2}x}}=\sec ^{2}x}$

${\displaystyle \cot ^{2}x+1={\frac {1}{\sin ^{2}x}}=\csc ^{2}x}$

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

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

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

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

Deze formules worden ook naar Carnot genoemd.

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

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:

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

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

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

${\displaystyle \operatorname {d} \!x={\frac {2\,dt}{1+t^{2}}}}$

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

Deze formules zijn naar Thomas Simpson genoemd.

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

Deling van de eerste door de tweede formule geeft

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

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

${\displaystyle \sin(x)+\sin(y)+\sin(z)-\sin(x+y+z)=4\sin {\dfrac {y+z}{2}}\sin {\dfrac {z+x}{2}}\sin {\dfrac {x+y}{2}}}$

${\displaystyle \cos(x)+\cos(y)+\cos(z)+\cos(x+y+z)=4\cos {\dfrac {y+z}{2}}\cos {\dfrac {z+x}{2}}\cos {\dfrac {x+y}{2}}}$