Question:
$\frac{\sin 3 x}{1+2 \cos 2 x}$ is equal to
(a) $\cos x$
(b) $\sin x$
(c) $-\cos x$
(d) $\sin x$
Solution:
(b) $\sin x$
We have,
$\frac{\sin 3 x}{1+2 \cos 2 x}=\frac{3 \sin x-4 \sin ^{3} x}{1+2\left(1-2 \sin ^{2} x\right)}$
$=\frac{3 \sin x-4 \sin ^{3} x}{1+2-4 \sin ^{2} x}$
$=\frac{\sin x\left(3-4 \sin ^{2} x\right)}{\left(3-4 \sin ^{2} x\right)}$
$=\sin x$
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