Question

Prove that:
$\cos {510}^{\circ }\cos {330}^{\circ }+\sin {390}^{\circ }\cos {120}^{\circ }=-1$

Answer 1

$\cos 510\cos 330+\sin 390\cos 120$

We have

$\cos 510=\cos \left(360+150\right)=\cos 150$

$=\cos \left(180-{30}^0\right)$

$=-\cos 30$

$\cos 330=\cos \left(360-30\right)=\cos 30$

$\sin 390=\sin \left(360+30\right)=\sin 30$

&$\cos 120=\cos \left(90+30\right)=-\sin {30}^0$

$\therefore \text{LHS}=-\cos 30\cdot \cos 30+\sin 3\theta \left(-\sin 30\right)$

$=-\left({\cos }^{2}30+{\sin }^{2}30\right)$

$=-1$

Hence proof