OnderwijsAnalyse – Goniometrie – Gelijkheden
\(\sin^2(\theta)+\cos^2(\theta)=1,\quad 1+\tan^2(\theta)=\dfrac{1}{\cos^2(\theta)},\quad 1+\dfrac{1}{\tan^2(\theta)}=\dfrac{1}{\sin^2(\theta)}\).
\(\sin(\theta+2\pi)=\sin(\theta),\quad\cos(\theta+2\pi)=\cos(\theta),\quad\tan(\theta+\pi)=\tan(\theta)\).
\(\sin(-\theta)=-\sin(\theta),\quad\cos(-\theta)=\cos(\theta),\quad\tan(-\theta)=-\tan(\theta)\).
\(\sin(\theta+\frac{1}{2}\pi)=\cos(\theta),\quad\cos(\theta+\frac{1}{2}\pi)=-\sin(\theta),\quad\sin(\pi-\theta)-\sin(\theta),\quad\cos(\pi-\theta)=-\cos(\theta),\quad\tan(\pi-\theta)=-\tan(\theta)\).
\(\sin(2x)=2\sin(x)\cos(x),\quad\cos(2x)=\cos^2(x)-\sin^2(x)=1-2\sin^2(x)=2\cos^2(x)-1\).
\(2\sin^2(x)=1-\cos(2x),\quad2\cos^2(x)=1+\cos(2x)\).
\(\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y),\quad\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)\).
\(\sin(x-y)=\sin(x)\cos(y)-\cos(x)\sin(y),\quad\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)\).
\(\tan(x+y)=\dfrac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)},\quad\tan(x-y)=\dfrac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\).
\(2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y),\quad2\cos(x)\cos(y)=\cos(x+y)+\cos(x-y),\quad2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y)\).
\[2\sin(x+y)\cos(x-y)=\sin(2x)+\sin(2y)\quad\text{en}\quad2\cos(x+y)\sin(x-y)=\sin(2x)-\sin(2y).\]
\begin{align*} \cos(x+y)\cos(x-y)&=\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\left(\cos(x)\cos(y)+\sin(x)\sin(y)\right) =\cos^2(x)\cos^2(y)-\sin^2(x)\sin^2(y)\\ &=\cos^2(x)\left(1-\sin^2(y)\right)-\left(1-\cos^2(x)\right)\sin^2(y) =\cos^2(x)-\sin^2(y). \end{align*} \begin{align*} \cos(x+y)\cos(x-y)&=\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\left(\cos(x)\cos(y)+\sin(x)\sin(y)\right) =\cos^2(x)\cos^2(y)-\sin^2(x)\sin^2(y)\\ &=\left(1-\sin^2(x)\right)\cos^2(y)-\sin^2(x)\left(1-\cos^2(y)\right) =\cos^2(y)-\sin^2(x). \end{align*} \begin{align*} \sin(x+y)\sin(x-y)&=\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\left(\sin(x)\cos(y)-\cos(x)\sin(y)\right) =\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)\\ &=\sin^2(x)\left(1-\sin^2(y)\right)-\left(1-\sin^2(x)\right)\sin^2(y) =\sin^2(x)-\sin^2(y). \end{align*} \begin{align*} \sin(x+y)\sin(x-y)&=\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\left(\sin(x)\cos(y)-\cos(x)\sin(y)\right) =\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)\\ &=\left(1-\cos^2(x)\right)\cos^2(y)-\cos^2(x)\left(1-\cos^2(y)\right) =\cos^2(y)-\cos^2(x). \end{align*}
Laatst gewijzigd op 6 augustus 2022
