If $(2^n+1)\theta=\pi$, $2^n\cos\theta \cos2\theta\cos2^2\theta \cdots\cos2^{n-1}\theta=$

Question

If $(2^n+1)\theta=\pi$, $2^n\cos\theta \cos2\theta \cos2^2\theta \cdots\cos2^{n-1}\theta=$

A)-1

B)1

C)1/2

D)none

If I take $\sin$ on both sides $$\sin(2^n+1)\theta=\sin\pi$$ Since $\sin\pi=0$ $$(2^n+1)\theta=0$$ $$\theta=0$$ and $$2^n+1=0$$ $$2^n=-1$$

So in the first equation, every value of cos will become 1 as $\cos0$ is always 1.

$2^n=-1$ So $(-1)1=-1$

But the answer is 1, so what is wrong with my process.

Answer 1

Use the fact that

$$2^n\cos\theta\cos2\theta\cos2^2\theta\cdots\cos2^{n-1}\theta = \frac{\sin2^n\theta}{\sin\theta}$$

As we have $(2^n+1)\theta = \pi \implies2^n\theta = \pi-\theta, \theta\ne0$

So,

$$2^n\cos\theta\cos2\theta\cos2^2\theta\cdots\cos2^{n-1}\theta=\frac{\sin2^n\theta}{\sin\theta} = \frac{\sin(\pi-\theta)}{\sin\theta}=\frac{\sin\theta}{\sin\theta}=1 \ $$