Question

If A, B, C, D be four angles whose sum is $\pi$ then prove that the sum of the products of their cosines taken two and two together is equal to the sum of the products of their sines takes similarly.

Answer 1

AB, AC, AD, BC, BD, CD. Six in number.
$A+B+C+D=\pi$
$\therefore A+B=\pi -(C+D)$ ..$\left(1\right)$
or $\cos (A+B)=-\cos (C+D)$
or $\cos A\cos B+\cos C\cos D=\sin A\sin B+\sin C\sin D$ ..$\left(2\right)$
Similarly write relations in AC, BD and AD; BC by taking
$A+C=\pi -(B+D)$
$A+D=\pi -(B+C)$
Adding these, we get
$\sum \cos A\cos B=\sum \sin A\sin B$
Six terms in each sigma.