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$\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } $

I am very much inquisitive to see how this trigonometrical identity can be proved.

PS:I am not much of interested about an inductive proof.

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    @Foool Then this is the generalized Morrie's law, if you insist. But I don't know why you would insist, since here the special case contains all the ideas of the general case, so no purpose is served by baking the distinction between them into our terminology.2012-03-11

2 Answers 2

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The proof is just repeated application of the double-angle formula for the sine function. For example, the case $n=3$: $\sin 8A = 2 \sin 4A \cos 4A = 2 (2 \sin 2A \cos 2A) \cos 4A = 2 (2 (2 \sin A \cos A) \cos 2A) \cos 4A.$

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    ...but Hans's solution is exactly what you'd be using for an inductive proof (note the word *repeated* in his answer).2010-10-31
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You can prove it by induction since

$\sin t = 2 \cos \dfrac{t}{2}\sin \dfrac{t}{2}$

$\sin t = 2^2 \cos \dfrac{t}{2}\cos \dfrac{t}{4}\sin \dfrac{t}{4}$

$\sin t = {2^3}\cos \dfrac{t}{2}\cos \dfrac{t}{4}\cos \dfrac{t}{8}\sin \dfrac{t}{8}$

So we conjecture:

$\sin t = 2^n\sin\dfrac{t}{2^n} \prod_{k=1}^{n} \cos\dfrac{t}{2^k} $

It is true for $n=1$

$\sin t = 2^1\sin\dfrac{t}{2^1} \prod_{k=1}^{1} \cos\dfrac{t}{2^1} = 2 \cos \dfrac{t}{2}\sin \dfrac{t}{2} $

But then for $n+1$ we get

$\sin t = 2^{n+1}\sin\dfrac{t}{2^{n+1}} \prod_{k=1}^{n+1} \cos\dfrac{t}{2^{k}} $

$\sin t = {2^n}2\sin \frac{t}{{{2^{n + 1}}}}\cos \frac{t}{{{2^{n + 1}}}}\prod\limits_{k = 1}^n {\cos } \frac{t}{{{2^k}}}$

$\sin t = {2^n}\sin \frac{t}{{{2^n}}}\prod\limits_{k = 1}^n {\cos } \frac{t}{{{2^k}}}$