Prove the following infinite trig product by induction.

Question

Prove by induction, that for a positive integrer $n$, that $$\cos x \times \cos2x \times \cos 4x \times \cos 8x ... \times\cos (2^nx) = \frac{\sin(2^{n+1}x)}{2^{n+1}\sin x}$$

So to start I'm gonna write $\cos x \times \cos2x \times \cos 4x \times \cos 8x ... \times\cos (2^nx)$ as $$\prod _{r=o} ^n \cos(2^r x)$$ I think this correct notation. Correct me if I'm wrong. So my proof so far is shown below, but I'm having trouble finishing it. Any help or suggestions to my work so far will be greatly appreciated. $$\prod _{r=o} ^n \cos(2^r x)= \frac{\sin(2^{n+1}x)}{2^{n+1}\sin x}$$ If $n=0$ then $$ LHS: \prod _{r=o} ^0 \cos(2^r x)= \cos x$$ $$RHS: \frac{\sin(2^{1}x)}{2^{1}\sin x} = \frac{2\sin x \cos x}{2\sin x} = \cos x $$ $\therefore$ RHS = LHS, so it is true when $n=o$

Assume true when $n=k$ $$\prod _{r=o} ^k \cos(2^r x)= \frac {\sin(2^{k+1}x)} {2^{k+1}\sin x}$$

If $n=k+1$

$$\prod _{r=o} ^{k+1} \cos(2^r x) = \left( \prod _{r=o} ^k \cos(2^r x) \right) \times \cos(2^{k+1}x)$$ $$\frac{\sin(2^{k+1}x)}{2^{k+1}\sin x} \times \cos(2^{k+1}x)$$

This is as far as I've got, any help would be appreciated.

Answer 1

You're really close. If our induction hypothesis is $$\prod_{r=0}^{k}\cos(2^r x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin x}$$ then $$\prod_{r=0}^{k+1}\cos(2^r x)=\prod_{r=0}^{k}\cos(2^r x)\cos(2^{k+1} x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin x} \cos(2^{k+1} x)$$ Now, I suggest you to use the expression of the sine of the double angle, that's to say, $$\sin(2y)=2\sin y\cos y$$ with $y=2^{k+1}x$.

Answer 2

You have to show that $$\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}\cdot \cos(2^{k+1}x)=\frac{\sin(2^{k+2}x)}{2^{k+2}\sin(x)}$$ and this is true since we have $$\sin(2^{k+1}x)\cos(2^{k+1}x)=\frac{1}{2}\sin(2^{k+2}x)$$