Question

Evaluate: $\cos {20}^{\circ }\cos {70}^{\circ }-\sin {20}^{\circ }\sin {70}^{\circ }$

Answer 1

$\cos 20°\cos 70°-\sin 70°\sin 20°=0$

$\Rightarrow \cos 20°\cos 70°-\sin 70°\sin 20°$

$=\cos (70°+20°)[\because \cos A\cos B-\sin A\sin B=\cos (A+B)]$

$=\cos 90°$

$=0$

Hence, the answer is solved.