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I inverted the LHS such that first term is $\cos(x/2^n)$ then and multiplied it by $2\sin(x/2^n)/2\sin(x/2^n)$ and evaluated it by $\sin2A=2\sin A\cos A$ but now I am stuck.Are any other ways to prove it? Please give suggestions.what should I do?

Bernard
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Science pills
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2 Answers2

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Prove by induction that:

$$ \prod_{i=1}^n\cos\frac{x}{2^i}=\frac{\sin x}{2^n\sin{\frac{x}{2^{n}}}}, $$

and let $n\rightarrow\infty$.

user
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Expand again and again $\sin(x)$ by $\sin(x)=2\cos(\frac{x}{2})\sin(\frac{x}{2})$ and so on, you have $$\frac{\sin x}{x} = \cos(\frac{x}{2})\cos(\frac{x}{4})\cos(\frac{x}{8})\cos(\frac{x}{16})\dots\cos(\frac{x}{2^n})\frac{\sin(\frac{x}{2^n})}{\frac{x}{2^n}}$$ and you let $n$ goes to infinity.

poyea
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