Simplifying Trig Product in terms of a single expression and $n$

Question

I participated in a math competition at a nearby university yesterday, and there was one ciphering problem that no one correctly simplified.

As the title suggests, we were asked to simplify the following trigonometric product such that it was in terms of a single expression (sine/cosine) and $n$.

$$\sin(x)\prod_{i=0}^n\cos(2^ix)$$

Typically with problems like these, which must be solved in under two minutes, the expression is telescoping in some manner that allows one to quickly simplify or solve it. This was how I approached it but simply didn't have enough time to see if it would work.

I know one of the professors, and he said the problem may be simplified using a number of double-angle formulas, but I'm still unsure how this method would work.

Could someone show how they might approach simplifying such an expression?

Answer 1

$$\sin(x)\prod_{i=0}^n\cos(2^ix)=\sin(x)\cos(x)\cos(2x)\cos(4x)...\cos(2^nx)= \\ =\frac{\sin(2x)}{2}\cos(2x)\cos(4x)...\cos(2^nx)= \\ =\frac{\sin(4x)}{4}\cos(4x)...\cos(2^nx)=...= \\ = \frac{\sin(2^{n+1}x)}{2^{n+1}}$$

Answer 2

Hint: We are looking at $$\sin x\cos x\cos 2x\cos 4x\cos 8x\cos 16 x\cdots.$$ This is $$\frac{1}{2}\cdot\sin 2x \cos 2x\cos 4x\cos 8x\cos 16x\cdots,$$ which is $$\frac{1}{4}\cdot\sin 4x \cos 4x\cos 8x\cos 16x\cdots.$$ Continue.