How to prove this trig identity?

Question

If $A+B+C=\pi$ then prove:$$\sin^2A+\sin^2B+\sin^2C=2-2\cos A\cos B\cos C$$

I am completely lost on this, please help.

Answer 1

$$\sin^2A+\sin^2B+\sin^2C=1-\cos^2C+1-(\cos^2A-\sin^2B)$$

Now, $\displaystyle\cos^2A-\sin^2B=\cos(A+B)\cos(A-B)=\cos(\pi-C)\cos(A-B)$ $=-\cos C\cos(A-B)$

So, $\displaystyle\cos^2C+\cos^2A-\sin^2B=\cos^2C-\cos C\cos(A-B)$

$\displaystyle=\cos C[\cos C-\cos(A-B)]$

$\displaystyle=\cos C[-\cos(A+B)-\cos(A-B)]$ as $A+B=\pi-C$

Now, $\displaystyle\cos(A+B)+\cos(A-B)=?$

Answer 2

$A+B+C = \pi \implies A + B = \pi - C \implies \sin(A+B)=\sin(C) \land \cos(A+B)=-\cos(C)$

$\sin^2(A) + \sin^2(B) + \sin^2(C) \\ = \sin^2(A) + 1-\cos^2(B) + 1 − \cos^2(A+B) \\ =2+(\sin^2(A)-\cos^2(B))-\cos^2(A+B) \\ = 2+\cos(B-A)\cos(A+B)-\cos(A+B)\cos(A+B) \\ = 2+(\cos(B-A)+\cos(A+B))\cos(A+B) \\ = 2 - 2\cos(A)\cos(B)\cos(C)$