Question

If $A+B+C+D=2\pi$, show that $\cos A-\cos B+\cos C-\cos D=4\sin \frac{A+B}{2}\sin \frac{A+D}{2}\cos \frac{A+C}{2}$.

Answer 1

$A+B+C+D=2\pi$
L.H.S. $=(\cos A+\cos C)-(\cos B+\cos D)$
$=2\cos \frac{A+C}{2}\cos \frac{A-C}{2}-2\cos \frac{B+D}{2}\cos \frac{B-D}{2}$
Now $\frac{B+D}{2}=\pi -\frac{A+C}{2}$,
$\therefore \cos \frac{B+D}{2}=-\cos \frac{A+C}{2}$
$\therefore$ L.H.S. $=2\cos \frac{A+C}{2}[\cos \frac{A-C}{2}+\cos \frac{B-D}{2}]$
$=2\cos \left(\frac{A+C}{2}\right)\left[2\cos \left(\frac{A+B-C-D}{4}\right)\cos \left(\frac{A+D-C-B}{4}\right)\right]$ $\left(1\right)$
Now $C+D=2\pi -(A+B),C+B=2\pi -(A+D)$
$\therefore \frac{(A+B)-(C-D)}{4}=\frac{(A+B)-2\pi +(A+B)}{4}$
$=\frac{A+B}{2}-\frac{\pi }{2}$
$\therefore \cos \frac{(A+B)-(C+D)}{4}=\cos (\frac{A+B}{2}-\frac{\pi }{2})=\sin \frac{A+B}{2}$ ..$\left(2\right)$
Again $\frac{(A+D)-(C+B)}{4}=\frac{(A+D)-2\pi +(A+D)}{4}=(\frac{A+D}{2}-\frac{\pi }{2})$
$\therefore \cos \frac{A+D-C-B}{4}=\cos (\frac{A+D}{2}-\frac{\pi }{2})=\sin \frac{A+D}{2}$ .$\left(3\right)$
Hence from $\left(1\right)$ with the help of $\left(2\right)$ and $\left(3\right)$, we get L.H.S. $=4\cos \frac{A+C}{2}\cdot \sin \frac{A+B}{2}\cdot \sin \frac{A+D}{2}$.