Mathematical Induction Angles proof.

Question

![this is a very dicy problem. It would be great to go into details of how to prove it using induction or any other alternate way is highly appreciated.][1]

$$\sin(x)\cos(x)\cos(2x)\cos(4x)\cos(8x)...\cos(2^nx)=\frac{\sin(2^{n+1}x)}{2^{n+1}}$$

this is a very dicy problem. It would be great to go into details of how to prove it using induction or any other alternate way is highly appreciated.

Answer 1

Try : $\sin(2a) = 2*\sin(a)*\cos(a)$

By induction: the latter proves the case n=0

Let that be true for n:

$sin(x)*cos(x)*...*cos(2^nx) = \frac{sin(2^{n+1}x)}{2^{n+1}}$

=> $sin(x)*cos(x)*...*cos(2^nx)*cos(2^{n+1}x) = \frac{sin(2^{n+1}x)}{2^{n+1}}*cos(2^{n+1}x) = \frac{1}{2^{n+1}}*\frac{sin(2^{n+2}x)}{2} $

Here I used the relation I gave above with $a= 2^{n+1}x$

=> $sin(x)*cos(x)*...*cos(2^nx)*cos(2^{n+1}x) = \frac{sin(2^{n+2}x)}{2^{n+2}} $

Answer 2

For the induction step: \begin{align*} [\sin(x)\cos(x) \cdots \cos(2^{n-1}x)] \cdot \cos(2^nx) &=\frac{\sin(2^{n}x)}{2^{n}} \cdot \cos(2^nx) &\text{induction hypothesis}\\ &=\frac{2\sin(2^{n}x)\cos(2^nx)}{2^{n+1}} &\text{multiply by $\frac{2}{2}$}\\ &=\frac{\sin(2^{n+1}x)}{2^{n+1}} &\text{double angle identity}\\ \end{align*}