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

this is just the $\cos$ rule and the $\sin$ rule combined 
$a^2 = b^2 + c^2 - 2bc \cos A$ and $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ where $R$ is the radius of the circumcircle of $ABC.$
the cosine rule becomes 
$\begin{align}
\sin^2 A  &= \sin^2 B + \sin^2 C-2\sin B \sin C \cos A \\
&=\cos^2 B + \cos^2 C - (\cos^2 B -\sin^2 B) -  (\cos^2 C -\sin^2 C) - 2\sin B \sin C \cos A\\
&=\cos^2 B + \cos^2 C - (\cos 2B + \cos 2C)- 2\sin B \sin C \cos A\\
&=\cos^2 B + \cos^2 C - 2\cos(B +C)\cos(B-C)- 2\sin B \sin C \cos A\\
&=\cos^2 B + \cos^2 C + 2\cos A (\cos B\cos C + \sin B \sin C)- 2\sin B \sin C \cos A\\
&=\cos^2 B + \cos^2 C + 2\cos A \cos B\cos C\\
\end{align}$
use of $\cos(B+C) = \cos(180^\circ - A) = -\cos A$ was used in the one before the last.
