# Trigonometry: Proving Question involving Sum to Product

I have a homework question that asks me to prove the following: $$\frac{\sin \theta+\sin 7\theta}{\sin 3\theta+\sin 5\theta}=2\cos2\theta-1$$ When I tried proving it, I could only do $$LHS=\frac{\sin \theta+\sin 7\theta}{\sin 3\theta+\sin 5\theta}$$ $$=\frac{2\sin4\theta\cos3\theta}{2\sin4\theta\cos\theta}$$ $$=\frac{\cos3\theta}{\cos\theta}$$

Is there a way to turn $\frac{\cos3\theta}{\cos\theta}$ into $2\cos2\theta-1$ or is there a different approach that I could take to the question? More importantly, I'd also like to know if I could simplify a cosine divided by a cosine like in this question here.

Edit: It's okay, I've found a solution that doesn't require the triple angle:

$$LHS=\frac{\cos3\theta}{\cos\theta}+1-1$$ $$=\frac{\cos3\theta+cos\theta}{\cos\theta}-1$$ $$=\frac{2cos2\theta cos\theta}{\cos\theta}-1$$ $$=2cos2\theta-1$$ $$=RHS$$

Thanks for all your help though guys.

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$$\frac{\cos 3x}{\cos x} = \frac{4 \cos^3 x-2\cos x}{\cos x} = 4\cos^2 x-3= 2(2\cos^2 x-1)-1= 2\cos 2x-1$$
$\cos3\theta=4\cos^3\theta -3\cos\theta$
$\cos2\theta=2\cos^2\theta-1$