Question

Find the value sin 330.cos 120+cos 210. sin 300
$\sin(330) \cos(120) + \cos(210) \sin(300)$

Answer 1

We know,

$\sin (330^o)=\sin (360^o-30^o)=\sin (-30^o)=-\sin (30^o)=-\frac{1}{2} \\ \cos (120^o)=\cos (180^o-60^o)=-\cos (60^o)=-\frac{1}{2} \\ \cos (210^o)=\cos (180^o+30^o)=-\cos (30^o)=-\frac{\sqrt{3}}{2} \\ \sin (300^o)=\sin (360^o-60^o)=\sin (-60^o)=-\sin (60^o)=-\frac{\sqrt{3}}{2}$

Plug in the above values in the given expression.

So the given expression is

$\sin (330^o)\cos (120^o)+\cos (210^o)\sin (300^o)=(-\frac{1}{2})^2 +(-\frac{\sqrt{3}}{2})^2\\ \sin (330^o)\cos (120^o)+\cos (210^o)\sin (300^o)=\frac{1+3}{4}=1$