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
 A: 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
$$\sin^2 A = \cos^2 B + \cos^2 C + 2 \cos A\cos B \cos C.$$
A: 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.
A: 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.
A: $$\cos^2B+\cos^2C+2\cos A\cos B\cos C =1+\cos^2B-\sin^2C+2\cos A\cos B\cos C$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $$\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$$
Again, $\cos(B+C)=\cos(\pi-A)=-\cos A$
Using Werner Formula,
$$2\cos A\cos B\cos C=\cos A[2\cos B\cos C]$$
$$=\cos A[\cos(B-C)+\cos(B+C)]=\cos A[\cos(B-C)-\cos A]$$
I hope the rest should not be too tough to deal with
