Trig multiple angle identity I have a quick problem that I can't solve. I know it's simple, but I can't seem to crack it. 
Determine whether the equation is an identity. Answer true or false. 
$$
\sin4\theta = \cos\theta(4\sin\theta-8\sin^3\theta)
$$
The answer is true. Any help would be appreciated. 
 A: You may write
$$
\begin{align}
\sin4\theta
&=2\sin(2\theta)\cos(2\theta)\\\\
&=\cos \theta \:(4\sin\theta\cos(2\theta))\\\\
&=\cos \theta \:(4\sin\theta(\cos^2\theta-\sin^2 \theta))\\\\
&=\cos \theta \:(4\sin\theta(1-2\sin^2 \theta))\\\\
&=\cos \theta \:(4\sin\theta-8\sin^3 \theta)\\\\
\end{align}
$$ as announced.
A: $$\sin(4x)= 2\sin(2x)\cos(2x) = 4\sin(x)\cos(x)(1 - 2\sin^2(x))$$
Therefore, $$\sin(4x) = \cos(x)(4\sin(x) - 8\sin^3(x))$$
A: Notice, $$RHS=\cos\theta(4\sin\theta-8\sin^3\theta)$$
$$=2\cos\theta(2\sin\theta-4\sin^3\theta)$$
$$=2\cos\theta((3\sin\theta-4\sin^3\theta)-\sin\theta)$$
$$=2\cos\theta(\sin3\theta-\sin\theta)$$
$$=2\sin3\theta\cos\theta-2\sin\theta\cos\theta$$
Applying $\color{red}{2\sin A\cos B=\sin(A+B)+\sin(A-B)}$
$$=\sin(3\theta+\theta)+\sin(3\theta-\theta)-\sin2\theta$$
$$=\sin4\theta+\sin2\theta-\sin2\theta$$
$$=\sin4\theta=LHS$$
Hence, the given equation: $\color{blue}{\sin4\theta=\cos\theta(4\sin\theta-8\sin^3\theta)}$  is an identity. 
A: HINT: Use the double angle identities:
$\sin{2\theta} = 2 \sin{\theta} \cos{\theta}$ and $\cos{2\theta} = 1-2 \sin^2{\theta}$.
From this point, I'm confident you can expand the left hand side of your problem to get the right hand side, but I'll leave you to it.
