Wiskunde Multiple Choice Vraag

sin 2α Als ——— = − 1 sin α tan 2α dan is ——— tan α gelijk aan	A. 1
B. 2
C. − 1
D. − 2
E. $\frac{8}{7}$

IN CONSTRUCTION	A.
B.
C.
D.
E.

1^ste manier :
$\\\frac{\sin2\alpha}{\sin\alpha}\!=\!-1\Leftrightarrow\frac{2.\sin\alpha\cos\alpha}{\sin\alpha}\!=\!-1\Leftrightarrow\cos\alpha\!=\!-\frac12\Leftrightarrow\cos120^\circ\Leftrightarrow\alpha\!=\!\pm 120^\circ+k.360^\circ$
$\\\frac{\tan 2{\alpha}}{\tan{\alpha}}=\frac{\tan(\pm240^\circ+k.720^\circ)}{\tan(\pm120^\circ+k.360^\circ)}=\frac{\tan(\pm240^\circ)}{\tan(\pm120^\circ)}=\frac{\tan 240^\circ}{\tan 120^\circ}=\frac{\tan 60^\circ}{-\tan 60^\circ}=-1$ 2^de manier :
$\\\frac{\sin2\alpha}{\sin\alpha}=-1\;\Leftrightarrow\;\frac{2.\sin\alpha\cos\alpha}{\sin\alpha}=-1\Leftrightarrow\cos\alpha=-\frac12$
$\cos2\alpha=2.\cos^2\alpha-1=2.\left(-\frac12\right)^2-1\;\Leftrightarrow\;\cos\alpha=-\frac12$
$\frac{\tan2\alpha}{\tan\alpha}=\frac{\sin2\alpha}{\sin\alpha}.\frac{\cos\alpha}{\cos2\alpha}=-1.\frac{-\frac12}{-\frac12}=-1$