Question

If $A,B,C,D$ are the angle of a quadrilateral, prove that $\cos \frac{1}{2}(A+B)+\cos \frac{1}{2}(C+D)=0$

Answer 1

We know that,

$\cos x+\cos y=2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$

Therefore,

$\cos \left(\frac{A+B}{2}\right)+\cos \left(\frac{C+D}{2}\right)=2\cos \left(\frac{A+B+C+D}{4}\right)\cos \left(\frac{A+B-C-D}{4}\right)$

$=2\cos \left(\frac{{360}^{\circ }}{4}\right)\cos \left(\frac{A+B-C-D}{4}\right)$ ( sum of the measures of all angles of a quadrilateral is 360 degree)

$=2\cos {90}^{\circ }.\cos \left(\frac{A+B-C-D}{4}\right)$

$=0$ ($\cos {90}^{\circ }=0$)