Question

What is the value of $\cos {20}^0+\cos {22}^0+\cos {158}^0+\cos {200}^0+\cos {300}^0$ ?

Answer 1

$\cos {158}^0=\cos ({180}^0-{22}^0)=-\cos {22}^0$
$\cos {200}^0=\cos ({180}^0+{20}^0)=-\cos {20}^0$
$\cos {300}^0=\cos ({360}^0-{60}^0)=\cos {60}^0=1/2$
$\therefore$ the given expression has the value
$=\cos {20}^0+\cos {22}^0-\cos {20}^0-\cos {22}^0+1/2=1/2$