Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$ if $A+B+C=180$ degrees I most humbly beseech help for this question.
If $A+B+C=180$ degrees, then prove
$$
\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)
$$
I am not sure what trig identity I should use to begin this problem. 

Let's take the Right Hand Side.
$$2 \sin B \cos C = \sin(B+C) + \sin(B-C) = \sin A + \sin(B-C)$$
Therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin A \sin(B-C)$
Now, $$2 \sin A \sin (B-C) = \cos (A - B +C) - \cos (A + B -C) = \cos 2C - \cos 2B = 2 \sin^2 B - 2 \sin ^2 C$$ Cancelling the $2$ gives us therefore, that $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$.
To finish, we already established $2 \sin A \sin B \cos C = \sin^2 A = \sin A \sin (B-C)$. Now, since $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$, therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin^2 B - \sin^2 C$.
