Question

$\sin \frac{13\pi }{3}.\sin \frac{2\pi }{3}+\cos \frac{4\pi }{3}.\sin \frac{13\pi }{6}=\frac{1}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

$\sin \left(\frac{13\pi }{3}\right)\sin \left(\frac{2\pi }{3}\right)+\cos \left(\frac{4\pi }{3}\right)\sin \left(\frac{13\pi }{6}\right)$

$=\sin (4\pi +\frac{\pi }{3})\sin (\pi -\frac{\pi }{3})+\cos (\pi +\frac{\pi }{3})\sin (2\pi +\frac{\pi }{6})$

$=\sin \frac{\pi }{3}\sin \frac{\pi }{3}-\cos \frac{\pi }{3}\sin \frac{\pi }{6}$

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

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