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. 

This may not be the shortest way, but is pretty systematic and doesn't involve any real trickery. First, we eliminate the $C$ angles from the equation, using that $C = 180 - A - B$. We write $\sin(C) = \sin(180 - A - B) = \sin(A + B)$, and $\cos(C) = \cos(180 - A - B) = -\cos(A + B)$ and what you need to prove is
$$\sin^2A + \sin^2B - \sin^2(A + B) = -2\sin A\sin B \cos(A+B)$$
Inserting the sine and cosine addition formulas in this, what you need to prove becomes
$$\sin^2A + \sin^2B - (\sin A\cos B + \cos A \sin B)^2 = -2\sin A \sin B(\cos A \cos B - \sin A \sin B)$$
Writing this out, it becomes
$$\sin^2A + \sin^2B - \sin^2 A\cos^2 B - 2\sin A \cos A \sin B \cos B -\cos^2 A \sin^2B$$
$$ = -2\sin A\sin B \cos A \cos B + 2\sin^2 A \sin^2 B$$
Cancelling terms, your goal is to prove 
$$\sin^2A + \sin^2B - \sin^2 A\cos^2 B-\cos^2 A \sin^2B = 2\sin^2 A \sin^2 B$$
Equivalently,
$$\sin^2A(1 - \cos^2B) + \sin^2B(1 - \cos^2A) = 2\sin^2 A \sin^2 B$$
This last equation is true since $1 - \cos^2 = \sin^2$. So going backwards in the above steps, one obtains the original equation $\sin^2A + \sin^2B - \sin^2C = 2\sin A \sin B \sin C$ as desired.
