Question

$\sin {120}^o\cos {150}^o+\cos {240}^o\sin {330}^o=?$

Answer 1

$\sin {120}^o\cos {150}^o+\cos {240}^o\sin {330}^o$

$\Rightarrow \sin ({180}^o-{60}^o)\cos ({180}^o-{30}^o)+\cos ({180}^o+{60}^o)\sin ({360}^o-{30}^o)$

$\Rightarrow -\sin {60}^o\cos {30}^o+(-\cos {60}^o)(-\sin {30}^o)$

$\Rightarrow \cos {60}^o\sin {30}^o-\sin {60}^o\cos {30}^o$

$\Rightarrow \sin ({30}^o-{60}^o)=\sin (-{30}^o)=-\sin {30}^o=\frac{-1}{2}$.