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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.

Harry Peter
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Quixotic
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  • Are you sure you have the right identity? It's not true for n=2. – J. M. ain't a mathematician Oct 31 '10 at 10:29
  • I do apologize for my mistake,I hope it is correct now. – Quixotic Oct 31 '10 at 10:35
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    I don't know how you will prove an $\mathbb N$ indexed family of identities without induction.. –  Oct 31 '10 at 11:19
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    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
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    This is Morrie's law. –  Mar 10 '12 at 17:29
  • @Steven Taschuk: Thanks for the update but looking at the wike page, I think Morrie's law is the special case of this identity. – Quixotic Mar 11 '12 at 07:47
  • @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. –  Mar 11 '12 at 15:00

3 Answers3

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

Hans Lundmark
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    I upvoted this; personally I'm still puzzled with the OP's aversion to induction. – J. M. ain't a mathematician Oct 31 '10 at 10:46
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    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. :) – Quixotic Oct 31 '10 at 10:49
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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). – J. M. ain't a mathematician Oct 31 '10 at 10:51
5

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

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

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

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

Pedro
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0

I tried to prove it formally, without direct mathematical induction.

$a_n=\cos {2^{n-1}A}$

$\Rightarrow a_n=\frac{2\sin {2^{n-1}A}\cdot\cos{2^{n-1}A}}{2\sin {2^{n-1}A}}$

$\Rightarrow a_n=\frac{\sin {2^{n}A}}{2\sin {2^{n-1}A}}$

$\Rightarrow a_n=\frac{V_{n+1}}{2V_n}$

Where, $V_n=\sin {2^{n-1}A}$

So,

$a_1\cdot a_2\cdot a_3\cdot a_4...a_n$

$=\frac{V_2}{2V_1}\cdot\frac{V_3}{2V_2}\cdot\frac{V_4}{2V_3}\cdot\frac{V_5}{2V_4}...\frac{V_{n+1}}{2V_n}$

$=\frac{V_{n+1}}{2^nV_1}$

$=\frac{\sin {2^nA}}{2^n\sin A}$

Note, here $2^{m-1}A≠z\pi$ ,where $m\in\Bbb N,\,n\in\Bbb N,\,z\in\Bbb Z$ and $\,m\le n$

P.S- I was also puzzled by the question of how to prove this identity without direct mathematical induction, then, I realised that I can use the same technique we use in finding some series by the Vn method.

Navneet
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