Question

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

• A. $\cos \theta$
• B. $\tan \theta$
• C. $\sin \theta$
• D. $\cos (-\theta )$

Answer 1

The correct option is C $\sin \theta$

Substituting the value of $\sin 3\theta$ and $\cos 2\theta$, we get

$=\frac{3\sin \theta -4{\sin }^3\theta }{1+2[2{\cos }^2\theta -1]}$

$=\frac{3\sin \theta -4{\sin }^3\theta }{1+4{\cos }^2\theta -2}$

$=\frac{3\sin \theta -4{\sin }^3\theta }{4{\cos }^2\theta -1}$

We know, ${\sin }^2\theta +{\cos }^2\theta =1$. We will replace 1 with ${\sin }^2\theta +{\cos }^2\theta$

So, we get,

$\frac{3\sin \theta -4{\sin }^3\theta }{4{\cos }^2\theta -{\sin }^2\theta -{\cos }^2\theta }$

$=\frac{3\sin \theta -4{\sin }^3\theta }{3{\cos }^2\theta -{\sin }^2\theta }$

$=\frac{3\sin \theta -4{\sin }^3\theta }{3(1-{\sin }^2\theta )-{\sin }^2\theta }$

$=\frac{\sin \theta (3-4{\sin }^2\theta )}{3-3{\sin }^2\theta -{\sin }^2\theta }$

$=\frac{\sin \theta (3-4{\sin }^2\theta )}{3-4{\sin }^2\theta }=\sin \theta$