Question

$\frac{1}{\sin 3\theta }[{\sin }^3\theta +{\sin }^3(\frac{2\pi }{3}+\theta )+{\sin }^3(\frac{4\pi }{3}+\theta )]$ equals

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $-\frac{3}{4}$
• D. none of these

Answer 1

The correct option is C $-\frac{3}{4}$

Using $\sin 3x=3\sin x-4{\sin }^{3}x\Rightarrow {\sin }^{3}x=\frac{3\sin x-\sin 3x}{4}$

$\frac{({\sin }^3\theta +{\sin }^3(\frac{2\pi }{3}+\theta )+{\sin }^3(\frac{4\pi }{3}+\theta ))}{\sin 3\theta }$$

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

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

$=\frac{(3\sin \theta +3(\sin (\frac{\pi }{3}+\theta )-\sin (\frac{\pi }{3}+\theta ))-3\sin 3\theta )}{4\sin 3\theta }$

$=\frac{(3\sin \theta +3\left(2\cos \frac{\pi }{3}\sin \theta \right)-3\sin 3\theta )}{4\sin 3\theta }$

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