Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\frac{\sin 8x}{\cos x.\cos 2x.\cos 4x.\sin x}$ is gelijk aan	A. 1
	B. 2
	C. 4
	D. 8
	E. 16

[ 5-7428 - op net sinds 19.8.2020-(E)-4.11.2023 ]

Translation in E N G L I S H

$https://latex.codecogs.com/png.image?\dpi{110}\frac{\sin 8x}{\cos x.\cos 2x.\cos 4x.\sin x}$ is equal to	A. 1
	B. 2
	C. 4
	D. 8
	E. 16

Oplossing - Solution

Gebruik herhaaldelijk de formule sin 2α = 2.sin α .cos α :
$\\\frac{\sin 8x}{\cos x.\cos 2x.\cos 4x.\sin x}=\frac{2\sin 4x\cos 4x}{\cos x.\cos 2x.\cos 4x.\sin x}=\frac{2.2\sin 2x.\cos 2x\cos 4x}{\cos x.\cos 2x.\cos 4x.\sin x}\\\\=\frac{2.2.2\sin x.\cos x.\cos 2x\cos 4x}{\cos x.\cos 2x.\cos 4x.\sin x}=8$