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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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2 Answers 2

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

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Welcome to math.SE. You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. –  Zev Chonoles Feb 22 '13 at 7:57

$\cos3\theta=4\cos^3\theta -3\cos\theta $

$\cos2\theta=2\cos^2\theta-1$

Use this.

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