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

Assume that $A$ is the largest angle (the modifications when this is not the case should be minor). Drop a perpendicular from $A$ to the side $BC$. Let $A'$ and $A''$ be the angles formed by dividing $A$ with this perpendicular, where $A'$ is adjacent to $B$ and $A''$ is adjacent to $C$. The trigonometry of right triangles states that $$\sin A' = \cos B,\ \sin A'' = \cos C,\ \cos A' = \sin B,\ \cos A'' = \sin C.$$ Since $A = A + A''$ (in terms of angle measure) you have
$$ \sin A  = \sin(A' + A'') = \sin A' \cos A'' + \cos A' \sin A'' = \cos B \sin C + \sin B \cos C,$$and thus  $$\sin^2 A = \cos^2 B \sin^2 C + 2 \sin B \cos B \sin C \cos C + \sin^2 B \cos^2 C.$$
Next write $\sin^2 B = 1 - \cos^2 B$ and $\sin^2 C = 1 - \cos^2 C$ to arrive at
\begin{align*}
\sin^2 A &= \cos^2 B + \cos^2 C - 2 \cos^2B \cos^2 C + 2 \sin B \cos B \sin C \cos C \\
&= \cos^2 B + \cos^2 C + 2 (\sin B \cos B \sin C \cos C - \cos^2B \cos^2 C) \\
&= \cos^2 B + \cos^2 C + 2 \cos B \cos C (\sin B \sin C - \cos B \cos C)
\end{align*}
Finally use the fact that
$$\cos A = \cos A' \cos A'' - \sin A \sin A'' = \sin B \sin C - \cos B \cos C$$
to get
$$\sin^2 A = \cos^2 B + \cos^2 C + 2 \cos A\cos B \cos C.$$
