Problem - 3902
In $\triangle{ABC}$, show that
\begin{align*}
&\sin^2A +\sin^2B+\sin^2C = 2 +2\cos A\cos B \cos C\\
&\cos^2A +\cos^2B + \cos^2C = 1-2\cos A\cos B\cos C
\end{align*}
The first equation can be derived from %%HREF%%3901%%.
\begin{align*}
&\sin^2 A +\sin^2 B + \sin^2 C\\
=\quad&\frac{1-\cos 2A}{2} + \frac{1-\cos 2B}{2}+\frac{1-\cos 2C}{2}\\
=\quad&\frac{1}{2}\cdot(3 - (\cos 2A + \cos 2B + \cos 2C))\\
=\quad&\frac{1}{2}\cdot(3 - (-1-4\cos A\cos B\cos C))\\
=\quad& 2 + 2\cos A\cos B \cos C
\end{align*}
Meanwhile,
\begin{align*}
&\cos^2 A + \cos^2 B + \cos^2 C\\
=\quad& 3 - (\sin^2 A + \sin^2 B + \sin^2 C)\\
=\quad& 3 - (2+2\cos A\cos B\cos C)\\
=\quad& 1 - 2\cos A \cos B \cos C
\end{align*}