Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\frac{\sin\alpha -2\sin^3\alpha }{2\cos^3\alpha -\cos\alpha }$ is gelijk aan	A. tan α
B. − tan α
C. cot α
D. − cot α
E. tan² α

$https://latex.codecogs.com/png.image?\dpi{110}\frac{\sin\alpha -2\sin^3\alpha }{2\cos^3\alpha -\cos\alpha }$ is equal to	A. tan α
B. − tan α
C. cot α
D. − cot α
E. tan² α

1^ste manier :
$Daar\;\left\{\begin{matrix}1-\cos2\alpha=2\sin^2\alpha\\1+\cos2\alpha=2\cos^2\alpha\\\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}1-\sin^2\alpha=\cos2\alpha\\2\cos^2\alpha-1=\cos2\alpha\\\end{matrix}\right.\;is$
$\frac{\sin{\alpha}-2\sin^3{\alpha}}{2\cos^3{\alpha}-\cos{\alpha}}=\frac{\sin{\alpha}\left(1-2\sin^2{\alpha}\right)}{\cos{\alpha}\left(2\cos^2{\alpha}-1\right)}=\frac{\sin{\alpha}}{\cos{\alpha}}=\tan{\alpha}$
2^de manier :
$\\\frac{\sin\alpha-2\sin^3\alpha}{2\cos^3\alpha-\cos\alpha}=\frac{\sin\alpha(1-2\sin^2\alpha)}{\cos\alpha(2\cos^2\alpha-)}\\=\frac{\sin\alpha(\sin^2\alpha+\cos^2\alpha-2\sin^2\alpha)}{\cos\alpha\left(2\cos^2\alpha-\sin^2\alpha-\cos^2\alpha\right)}\\=\frac{\sin{\alpha}\left(\cos^2{\alpha}-\sin^2{\alpha}\right)}{\cos{\alpha}\left(\cos^2{\alpha}-\sin^2{\alpha}\right)}=\frac{\sin{\alpha}}{\cos{\alpha}}=\tan{\alpha}$
$\\\frac{\sin\alpha-2\sin^3\alpha}{2\cos^3\alpha-\cos\alpha}=\frac{\sin\alpha(1-2\sin^2\alpha)}{\cos\alpha(2\cos^2\alpha-)}\\=\frac{\sin\alpha(\sin^2\alpha+\cos^2\alpha-2\sin^2\alpha)}{\cos\alpha\left(2\cos^2\alpha-\sin^2\alpha-\cos^2\alpha\right)}\\=\frac{\sin{\alpha}\left(\cos^2{\alpha}-\sin^2{\alpha}\right)}{\cos{\alpha}\left(\cos^2{\alpha}-\sin^2{\alpha}\right)}=\frac{\sin{\alpha}}{\cos{\alpha}}=\tan{\alpha}$