Question

Prove that : $sin(-{330}^{\circ })cos(-{300}^{\circ })+sin(-{420}^{\circ })cos{390}^{\circ }=-\frac{1}{2}$

Answer 1

$\sin (-{330}^o)=-\sin {330}^o$

$=-\sin ({360}^o-{30}^o)$

$=-(-\sin {30}^o)$

$=\frac{1}{2}$

$\cos (-{300}^o)=\cos {300}^o$

$=\cos ({360}^o-{60}^o)$

$=\cos {60}^o$

$=\frac{1}{2}$

$\sin (-{420}^o)=-\sin {420}^o$

$=-\sin ({360}^o+{60}^o)$

$=-\sin {60}^o$

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

$\cos {390}^o=\cos ({360}^o+{30}^o)$

$=\cos {30}^o$

$=\frac{\sqrt{3}}{2}$.

$\therefore \sin (-{330}^o)\cos (-{300}^o)+\sin (-{420}^o)\cos {390}^o$

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

$=-1/2$.