Question

If $A+B+C=180$, Prove that ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}c=2+2\cos A.\cos B.\cos c$

Answer 1

Consider the problem,

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}c=2+2\cos A.\cos B.\cos c$

We can write ${\sin }^{2}A$ as,

${\sin }^{2}A=\frac{1-\cos \left(2A\right)}{2}$

Therefore,

$\begin{array}{c}LHS=\frac{1-\cos \left(2A\right)}{2}+\frac{1-\cos \left(2B\right)}{2}+\frac{1-\cos \left(2C\right)}{2}\\ =\frac{3}{2}-\left(\cos \left(2A\right)+\cos \left(2B\right)+\cos \left(2C\right)\right)\\ =\frac{1}{2}\left(3-\left(2\cos \left(A+B\right)\cos \left(A-B\right)+\cos \left(2C\right)\right)\right)\\ C=180-\left(A+B\right)\\ \cos \left(C\right)=\cos \left(180-\left(A+B\right)\right)\\ \cos \left(C\right)=-\cos \left(A+B\right)\end{array}$

Therefore,

$\begin{array}{c}=\frac{3}{2}-\left(-2\cos \left(C\right)\cos \left(A-B\right)+\cos \left(2C\right)\right)\\ \cos \left(2C\right)=2{\cos }^2\left(C\right)-1\end{array}$

And,

$\begin{array}{c}=\frac{1}{2}\left(3-\left(-2\cos C\right)\cos \left(A-B\right)+2{\cos }^2\left(C\right)-1\right)\\ =\frac{1}{2}\left(4-\left(2\cos \left(C\right)\cos \left(C\right)-\cos \left(A-B\right)\right)\right)\\ =\frac{1}{2}\left(4-2\cos \left(C\right)\left(-\cos \left(A+B\right)-\cos \left(A-B\right)\right)\right)\\ =\frac{1}{2}\left(4+2\cos \left(C\right)\left(\cos \left(A+B\right)\cos \left(A-B\right)\right)\right)\\ =\frac{1}{2}\left(4+2\cos \left(C\right)\times 2\cos \left(A\right)\cos \left(B\right)\right)\\ =2+2\cos \left(A\right)\cos \left(B\right)\cos \left(C\right)\end{array}$

Therefore, If $A+B+C=180$, ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}c=2+2\cos A.\cos B.\cos c$