Question

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

• A. cos$\theta$
• B. cos$(\frac{\pi }{2}-\theta )$
• C. sin$\theta$
• D. cos$(-\theta )$

Answer 1

Answer: (B), (C)

sin$\theta$

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

Substituting the value of $\sin 3\theta \&\cos 2\theta$

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

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

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

Further, replace 1 with ${\sin }^2\theta +{\cos }^2\theta$

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

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