trigonometry - If $A+B+C=π$, prove that

If $A+B+C=π$, prove that
$$\cos^2A+\cos^2B-\cos^2C=-2\cos A\cdot\cos B\cdot\cos C.$$

ATTEMPT:
Given $$A+B+C=π,$$
$$A+B=π-C$$
Taking "cos" on both sides

$$\cos(A+B)=-\cos C.$$

Now,
\begin{align*}
L.H.S&=\cos^2A+\cos^2B-\cos^2C\\
&=\frac{1+\cos 2A}{2}+\frac{1+\cos 2B}{2}-\cos^2 C\\
&=\frac{2+\cos 2A+\cos 2B}{2}-\cos^2C.
\end{align*}

How should I complete now?

Answer

The identity

$$1 - \cos^2 A - \cos^2 B - \cos^2 C - 2 \cos A \cos B \cos C = 0$$

can be seen as a re-writing of the relation $\cos C = - \cos(A+B)$ without sines of $A$ and $B$:

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

$$\cos C + \cos A \cos B = \sin A \sin B$$
$$\left(\;\cos C + \cos A \cos B\;\right)^2 = \sin^2 A \sin^2 B = \left(1-\cos^2 A\right)\left(1-\cos^2 B\right)$$
$$\cos^2 C + 2 \cos A \cos B \cos C \color{red}{+ \cos^2A \cos^2 B} = 1 - \cos^2 A - \cos^2 B \color{red}{+ \cos^2 A \cos^2 B}$$
and the result follows.