Question

If $A+B+C=2S$, show that ${\cos }^{2}S+{\cos }^2(S-A)+{\cos }^2(S-B)+co{s}^2(S-C)=2+2\cos A\cos B\cos C$.

Answer 1

Change in terms of double angles, i.e.,
${\cos }^{2}A=\frac{1}{2}(1+\cos 2A)$
Also $A=B+C=2S$
L.H.S. $=\frac{1}{2}[1+\cos 2S+1+\cos (2S-2A)+1+\cos (2S-2B)+1+\cos (2S-2C\left)\right]$.
$=2+\frac{1}{2}\left[2\cos \right(2S-A)\cos A+2\cos (2S-B-C\left)\cos \right(B-C\left)\right]$
$=2+\cos A\left[\cos \right(B+C)+\cos (B-C\left)\right],\because 2S-A=B+C$
$=2+\cos A\left(2\cos B\cos C\right)$
$=2+2\cos A\cos B\cos C$.
Alternative:
$\therefore L.H.S.=1-{\sin }^{2}S+1-{\sin }^2(S-A)+{\cos }^2(S-B)+{\cos }^2(S-C)$
$=2+\left[{\cos }^2\right(S-B)-{\sin }^2]+\left[{\cos }^2\right(S-C)-{\sin }^2(S-A\left)\right]$
Now apply
${\cos }^{2}A-{\sin }^{2}B=\cos (A+B)\cos (A-B)$ etc.