# Proving trigo identity

Prove (given that A+B+C=π):
$$\sin^2 A + \sin^2 B - \sin^2 C = 2 \sin A \sin B \sin C$$
I got this far:
$$\begin{matrix}\mbox{LHS} & = & \sin^2 A + \sin^2 B - \sin^2 C \\ \ & = & \sin^2 A + \left ( \sin B + \sin C \right ) \left ( \sin B - \sin C \right ) \\ \ & = & \sin^2 A + \left ( 2 \sin \frac{B+C}{2} \cos \frac{B-C}{2} \right) \left ( 2 \cos \frac{B+C}{2} \sin \frac{B-C}{2} \right ) \\ \ & = & \sin^2 A + \left (2 \sin \frac{A}{2} \cos \frac{B-C}{2} \right) \left ( 2 \cos \frac{-A}{2} \sin \frac{B-C}{2} \right ) \\ \ & = & \sin^2 A + \left ( -2 \sin \frac{A}{2} \cos \frac{A}{2} \right ) \left ( 2 \cos \frac{B-C}{2} \sin \frac{B-C}{2} \right ) \\ \ & = & \sin^2 A + \left ( -2 \sin A \right ) \left (2 \cos \frac{B-C}{2} \sin \frac{B-C}{2} \right ) \end{matrix}$$

Help please...

TIA

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## Answers and Replies

In its current form, it's false. Try a = b = c = pi/3.

Never mind, I got it already... right hand side should be cos C instead of sin C.