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.
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