# Proving: $\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A … \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A }$

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

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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Are you sure you have the right identity? It's not true for n=2. – J. M. Oct 31 '10 at 10:29
I do apologize for my mistake,I hope it is correct now. – VelvetThunder Oct 31 '10 at 10:35
I don't know how you will prove an $\mathbb N$ indexed family of identities without induction.. – anon Oct 31 '10 at 11:19
Perhaps one could use complex analysis, using Picard's theorem or something like that. But I can't seem to get the argument right. So you can easily check that the ratio of the left-hand side and the right-hand side is an entire function with no zeros. But that is as far as I can get... – Braindead Oct 31 '10 at 17:44
This is Morrie's law. – user21467 Mar 10 '12 at 17:29

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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I upvoted this; personally I'm still puzzled with the OP's aversion to induction. – J. M. Oct 31 '10 at 10:46
It's not aversion,but I believe that I can prove it by induction myself, here I am more interested to see something tricky ! Now why I am asking for something tricky since I have noticed that once I see something like that it becomes easier for me to remember the identity, I hope my point is clear. :) – VelvetThunder Oct 31 '10 at 10:49
...but Hans's solution is exactly what you'd be using for an inductive proof (note the word repeated in his answer). – J. M. Oct 31 '10 at 10:51

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

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

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

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

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