Question

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

Answer 1

$$\frac{13\pi }{3}={780}^{\circ },\frac{2\pi }{3}={120}^{\circ },\frac{4\pi }{3}={240}^{\circ },\frac{13\pi }{6}={390}^{\circ }$$
LHS = $$\sin \left({780}^{\circ }\right)\sin \left({120}^{\circ }\right)+\cos \left({240}^{\circ }\right)\sin \left({390}^{\circ }\right)$$
$$=\sin ({90}^{\circ }\times 8+{60}^{\circ })\sin ({90}^{\circ }\times 1+{30}^{\circ })+\cos ({90}^{\circ }\times 2+{60}^{\circ })\sin ({90}^{\circ }\times 4+{30}^{\circ })$$
$$=\sin {60}^{\circ }\cos {30}^{\circ }+[-\cos {60}^{\circ }]\sin {30}^{\circ }$$
$$=\sin {60}^{\circ }\cos {30}^{\circ }-\cos {60}^{\circ }\sin {30}^{\circ }$$
$$=\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}$$
$$=\frac{3}{4}-\frac{1}{4}$$
$$=\frac{1}{2}$$
 = RHS
Hence proved.