Problem - 3900
In $\triangle{ABC}$, show that
\begin{align*}
\sin 2A + \sin 2B + \sin 2C &= 4\sin A\sin B \sin C\\
\cos 2A + \cos 2B + \cos 2C &= -1-4\cos A\cos B\cos C
\end{align*}
Both identities can be proved by applying basic formulas.
\begin{align*}
&\sin 2A + \sin 2B + \sin 2C \\
&= 2\sin(A+B)\cos (A-B) + 2\sin C\cos C\\
&= 2\sin C \cos (A-B) + 2\sin C\cos C\\
&= 2\sin C (\cos (A-B) - \cos (A+B)\\
&= 2\sin C \cdot 2 \sin A \sin B\\
&= 4\sin A \sin B \sin C
\end{align*}
\begin{align*}
&\cos 2A + \cos 2B + \cos 2C \\
&= 2\cos(A+B)\cos (A-B) + 2\cos^2C - 1\\
&= -2\cos C \cos (A-B) - 2\cos C\cos(A+B) -1\\
&= -2\cos C(\cos (A-B) + \cos (A+B) -1\\
&= -4\cos A\cos B\cos C -1
\end{align*}