Verify this identity: $\sin6\alpha + \sin2\alpha = \frac{\cos^22\alpha-\cos2\alpha\cos6\alpha}{\sin2\alpha}$ Say I want to verify the following identity:
$$\sin6\alpha + \sin2\alpha = \frac{\cos^22\alpha-\cos2\alpha\cos6\alpha}{\sin2\alpha}$$
I'm working on the first member of the identity so that it will eventually match the second one.
$$\sin6\alpha + \sin2\alpha =$$ 
$$\sin(2\times3\alpha)+\sin2\alpha=$$
$$2\sin3\alpha\cos3\alpha+2\sin\alpha\cos\alpha$$
That doesn't to be taking me anywhere. Any hints?
 A: The identity is equivalent (for $\sin2\alpha\ne0$) to
$$
\sin^22\alpha+\sin2\alpha\sin6\alpha-\cos^22\alpha+\cos2\alpha\cos6\alpha=0
$$
Note that
$$
\cos2\alpha\cos6\alpha+\sin2\alpha\sin6\alpha=\cos(6\alpha-2\alpha)
=\cos4\alpha
$$
and
$$
\cos^22\alpha-\sin^22\alpha=\cos4\alpha
$$
A: Use the triple angle formulas:
$$\cos 6\alpha = 4 \cos^3 2\alpha - 3\cos 2\alpha$$
$$\sin 6\alpha = -4 \sin^3 2\alpha + 3\sin 2\alpha$$
A: $$\frac{\cos^22\alpha-\cos2\alpha\cos6\alpha}{\sin2\alpha}=\frac{\cos 2\alpha(\cos2\alpha-\cos6\alpha)}{\sin2\alpha}\\
\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,=\frac{2\cos 2\alpha(\sin2\alpha\times\sin 4\alpha)}{\sin2\alpha}\\
\qquad\qquad\qquad\,=2\cos 2\alpha\sin 4\alpha\\
\\
\\
\,\,\,\qquad\qquad\qquad=\sin 6\alpha+\sin 2\alpha
$$
For more details, Check the list of trigonometric identities
A: There's a general fact that can be useful in these problems: Expressed as functions of $\cos\theta$, the trigonometric expressions $\cos n\theta$ and $\frac{\sin {(n\theta+\theta)}}{\sin \theta}$ can be expressed as polynomials. (Specifically, they'll be the $n$th Chebyshev polynomials of the first and second kind respectively.)
With this in mind, if we divide both sides of the desired identity by $\sin2\alpha$, we have
$$\frac{\sin6\alpha}{\sin 2\alpha} + 1  = \frac{\cos^22\alpha-\cos2\alpha\cos6\alpha}{\sin^2 2\alpha} = \frac{\cos 2\alpha-\cos6\alpha}{1-\cos^2 2\alpha}\cos2\alpha.$$ But the above principle indicates we can express both sides in terms of $\cos 2\alpha$, and indeed
\begin{align}
\frac{\sin6\alpha}{\sin 2\alpha}+1
&=(4\cos^2 2\alpha-1)+1\\
&=4\cos^2 2\alpha,\\
\frac{\cos 2\alpha-\cos6\alpha}{1-\cos^2 2\alpha}\cos2\alpha
&=\frac{\cos 2\alpha-(4\cos^3 2\alpha-3 \cos2\alpha )}{1-\cos^2 2\alpha}\cos2\alpha\\
&=\frac{4\cos 2\alpha(1-\cos^2 2\alpha)}{1-\cos^2 2\alpha}\cos 2\alpha\\
&=4\cos^2 2\alpha.
\end{align}
(In each case the first equality amounts to a triple angle identity). So the two sides indeed agree.
