Question

prove that: ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2+2\cos A\cos B\cos C$.

Answer 1

On changing ${\sin }^{2}A$ to $1-{\cos }^{2}A$ etc.

Proceeding directly as we need $2$, we write ${\sin }^{2}A=1-{\cos }^{2}A$ and

${\sin }^{2}B=1-{\cos }^{2}B$

$L.H.S.=2-{\cos }^{2}A-{\cos }^{2}B+{\sin }^{2}C$

$=2-({\cos }^{2}A-{\sin }^{2}C)-co{s}^{2}B$

$=2-\cos (A+C)\cos (A-C)-{\cos }^{2}B$

$=2+\cos B\cos (A-C)-{\cos }^{2}B$

$=2+\cos B\left[\cos \right(A-C)-\cos B]$

$=2+\cos B\left[\cos \right(A-C)+\cos (A+C\left)\right]$

$=2+\cos B\left(2\cos A\cos C\right)$

$=2+2\cos A\cos B\cos C$.