Question

$\frac{\sin 3\mathrm{\theta }}{1+2\cos 2\mathrm{\theta }}=$

(a) $\cos \mathrm{\theta }$
(b) $\sin \mathrm{\theta }$
(c) $-\cos \mathrm{\theta }$
(d) $\sin \mathrm{\theta }$

Answer 1

(b) $\sin \mathrm{\theta }$

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