Proving $\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$ As the title,
By considering $\bigtriangleup$ABC, Prove
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$$
Thanks

Since $A+B+C=\pi$, we can write
$$
\sin^2 A=\sin^2(\pi-(B+C))=\sin^2(B+C)
$$
Expand with the sum formula:
$$
\sin^2B\cos^2C+2\sin B\sin C\cos B\cos C+\cos^2B\sin^2C
$$
Transform $\sin^2$ into $\cos^2$:
$$
\cos^2C-\cos^2B\cos^2C+2\sin B\sin C\cos B\cos C+\cos^2B-\cos^2B\cos^2C
$$
Reduce and collect:
$$
\cos^2B+\cos^2C-2\cos B\cos C(\cos B\cos C-\sin B\sin C)
$$
Finish up.
