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 } $

Question

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

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.

Answer 1

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

Answer 2

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

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

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

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

Answer 3

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.