Question

Simplify:
$\cos {100}^{\circ }\cos {40}^{\circ }+\sin {100}^{\circ }\sin {40}^{\circ }$

Answer 1

Given expression:

$\cos {100}^{\circ }\cos {40}^{\circ }+\sin {100}^{\circ }\sin {40}^{\circ }$

This is in the form of,

$\cos A\cos B+\sin A\sin B=\cos (A-B)$

$\Rightarrow \cos {100}^{\circ }\cos {40}^{\circ }+\sin {100}^{\circ }\sin {40}^{\circ }$

$=\cos ({100}^{\circ }-{40}^{\circ })$

$=\cos {60}^{\circ }$

$=\frac{1}{2}$

$\therefore \cos {100}^{\circ }\cos {40}^{\circ }+\sin {100}^{\circ }\sin {40}^{\circ }=\frac{1}{2}$