Question

$\frac{\sin 3x}{1+2\cos 2x}$ is equal to

(a) cos x
(b) sin x
(c) – cos x
(d) sin x

Answer 1

(b) sin x

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{\sin 3x}{1+2\cos 2x}=\frac{3\sin x-4{\sin }^{3}x}{1+2\left(1-2{\sin }^{2}x\right)} \\ =\frac{3\sin x-4{\sin }^{3}x}{1+2-4{\sin }^{2}x} \\ =\frac{\sin x\left(3-4{\sin }^{2}x\right)}{\left(3-4{\sin }^{2}x\right)} \\ =\sin x \end{gathered}$$