Question

In any quadrilateral $ABCD$, prove that
(i) $\sin (A+B)+\sin (C+D)=0$
(ii) $\cos (A+B)=\cos (C+D)$.

Answer 1

In a quadrilateral $ABCD$, $\angle A+\angle B+\angle C+\angle D=360$

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

$=2\sin \left(180\right).\cos \left(\frac{A+B-C-D}{2}\right)$

$=0$

(ii) $\cos (A+B)-\cos (C+D)=-2\sin \left(\frac{A+B+C+D}{2}\right).\sin \left(\frac{A+B-C-D}{2}\right)=0$