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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?

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  • $\begingroup$ Try to simplify from RHS rather than LHS $\endgroup$ – Zack Ni Aug 26 '16 at 19:30
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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 $$ \cos^22\alpha-\sin^22\alpha=\cos4\alpha $$

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  • $\begingroup$ Wow, that makes it really easy! $\endgroup$ – John Aug 26 '16 at 20:08
  • $\begingroup$ Always fun, those trig identities...+1 $\endgroup$ – imranfat Aug 26 '16 at 21:28
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Use the triple angle formulas:

$$\cos 6\alpha = 4 \cos^3 2\alpha - 3\cos 2\alpha$$ $$\sin 6\alpha = -4 \sin^3 2\alpha + 3\sin 2\alpha$$

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$$\frac{\cos^22\alpha-\cos2\alpha\cos6\alpha}{\sin2\alpha}=\frac{\cos 2\alpha(\cos2\alpha-\cos6\alpha)}{\sin2\alpha}\\ \qquad\qquad\qquad\qquad\qquad\,\,\,\,\,=\frac{2\cos 2\alpha(\sin2\alpha\times\sin 4\alpha)}{\sin2\alpha}\\ \qquad\qquad\qquad\,=2\cos 2\alpha\sin 4\alpha\\ \\ \\ \,\,\,\qquad\qquad\qquad=\sin 6\alpha+\sin 2\alpha $$ For more details, Check the list of trigonometric identities

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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.

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  • $\begingroup$ I should hasten to add that the above strategy doesn't have the virtue of efficiency. The main advantage it has is that, once one has written things in terms of $\cos 2\alpha$, one is working with polynomials and rational functions which are easier to simplify and rearrange. $\endgroup$ – Semiclassical Aug 26 '16 at 20:17

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