Question

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

Answer 1

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

$=\cos (2\times 360-210)\cos (360-30)+\sin (360+30)\sin (180-60)$

$=(-\cos 30)\left(\cos 30\right)+\left(\sin 30\right)(-\sin 30)$

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

$=-1$

Hence proved.