Question

Prove that
$\sin {40}^{\circ }\cos {390}^{\circ }+\cos (-{300}^{\circ })\sin (-{330}^{\circ })=1$.

Answer 1

$\sin {420}^{\circ }\cos {390}^{\circ }+\cos (-{300}^{\circ })\sin (-{330}^{\circ })$
$=\sin ({360}^{\circ }+{60}^{\circ })\cos ({360}^{\circ }+{30}^{\circ })-\cos ({270}^{\circ }+{30}^{\circ })\sin ({270}^{\circ }+{60}^{\circ })$
$=\sin {60}^{\circ }\cos {30}^{\circ }+\sin {30}^{\circ }\cos {60}^{\circ }$
$=\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}$
$=\frac{3}{4}+\frac{1}{4}=1$