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I'm supposed to prove by induction the equality $$\prod_{k=1}^n\cos\frac{x}{2^k}=\frac{\sin x}{2^n\sin\frac{x}{2^n}}$$ I've shown the base case for $n=1$, assumed it is valid for $n$ then tried extending for $n+1$. I've used trigonometric identities to obtain the expression $$\left(2-2\cos\frac{x}{2^n}\right)\cos^2\frac{x}{2^k}=\sin ^2\frac{x}{2^k}$$ and it seems to me that I'm close, but I'm stuck here. Any help please?

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    $\begingroup$ I think you're on the wrong track. Instead, simply use the double-angle formula $\sin 2\theta = 2 \sin\theta \cos\theta$, taking $\theta = x/2^{n+1}$, to re-write the denominator. $\endgroup$
    – Blue
    Nov 22 '16 at 17:09
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To get the proof consider the identity $\cos(x)\sin(x) = \frac{1}{2}\sin(2x)$. Now the induction step leads to $$ \prod^{n+1}_{k=1} \cos(x/2^k) = \cos(x/2^{n+1}) \prod^{n}_{k=1} \cos(x/2^k). $$ On the last term you use the induction hypothesis. Using the proposed equality the first term is equal to $$ \cos(x/2^{n+1}) = \frac{\sin(x/2^n)}{2\sin(x/2^{n+1})} $$ Simplifying should yield the result.

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Hint: $\cos{x} \cos{\frac{x}{2}} = \frac{\cos x \cos{\frac{x}{2}} \sin{\frac{x}{2}}}{\sin{\frac{x}{2}}}$

and

$\sin{2x} = 2\sin{x}\cos{x}$

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  • $\begingroup$ I accepted the other answer simply because that user has less reputation than you :) $\endgroup$
    – bonehead
    Nov 23 '16 at 10:08

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