Question

Prove that:
$\sin \frac{13\pi }{3}\sin \frac{8\pi }{3}+\cos \frac{2\pi }{3}\sin \frac{5\pi }{6}=\frac{1}{2}$

Answer 1

we have to prove:

$\sin \frac{13\pi }{3}\sin \frac{8\pi }{3}+\cos \frac{2\pi }{3}\sin \frac{5\pi }{6}=\frac{1}{2}$

$L.H.S=\sin \frac{13\pi }{3}\sin \frac{8\pi }{3}+\cos \frac{2\pi }{3}\sin \frac{5\pi }{6}$

$=\sin (4\pi +\frac{\pi }{3})\sin (3\pi -\frac{\pi }{3})+\cos (\pi -\frac{\pi }{3})\sin (\pi -\frac{\pi }{6})$

$\because \sin (2n\pi +x)=\sin x,\cos (\pi -x)=-\cos x$

$\sin (n\pi -x)=\sin x,n=1,2,\mathrm{3.....}$

$\Rightarrow LHS=\sin \frac{\pi }{3}\sin \frac{\pi }{3}+(-\cos \frac{\pi }{3})\sin \frac{\pi }{6}$

$=\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}-\frac{1}{4}$

$=\frac{1}{2}=RHS$.

$\therefore \sin \frac{13\pi }{3}\sin \frac{8\pi }{3}+\cos \frac{2\pi }{3}\sin \frac{5\pi }{6}=\frac{1}{2}$

Hence proved.