Question

For all $N$, $\cos \left(\theta \right)·\cos \left(2\theta \right)·\cos \left(4\theta \right)·...·\cos \left(2^{n-1}\theta \right)=\_\_\_\_\_$

Answer 1

Solve for the required value

Given, $\cos \left(\theta \right)·\cos \left(2\theta \right)·\cos \left(4\theta \right)·...·\cos \left(2^{n-1}\theta \right)=\_\_\_\_\_$

On multiplying numerator and denominator by $2\sin \theta$, we get,

$$\begin{gathered} \cos \left(\theta \right)·\cos \left(2\theta \right)·\cos \left(4\theta \right)·...·\cos \left(2^{n-1}\theta \right) \\ =\frac{2\sin \left(\theta \right)\cos \left(\theta \right)·\cos \left(2\theta \right)·\cos \left(4\theta \right)·...·\cos \left(2^{n-1}\theta \right)}{2\sin \theta } \\ =\frac{\sin \left(2\theta \right).\cos \left(2\theta \right).\cos \left(4\theta \right)·...·\cos \left(2^{n-1}\theta \right)}{2\sin \left(\theta \right)}\left[\because 2\sin \left(\theta \right)\cos \left(\theta \right)=\sin \left(2\theta \right)\right] \end{gathered}$$

$=\frac{2\sin \left(2\theta \right)·\cos \left(2\theta \right)·\cos \left(4\theta \right)·...·\cos \left(2^{n-1}\theta \right)}{2^2\sin \left(\theta \right)}$ [By multiplying numerator and denominator by $2$]

$$\begin{gathered} =\frac{\sin \left(4\theta \right)·\cos \left(4\theta \right)·...·\cos \left(2^{n-1}\theta \right)}{2\sin \left(\theta \right)} \\ =\frac{2\sin \left(4\theta \right)·\cos \left(4\theta \right)·...·\cos \left(2^{n-1}\theta \right)}{2^3\sin \left(\theta \right)} \\ ⋮ \end{gathered}$$

$=\frac{2\sin \left(2^{n-1}\theta \right)·\cos \left(2^{n-1}\theta \right)}{2^n\sin \left(\theta \right)}$ [By repeating $n$ times]

$=\frac{\sin \left(2^n\theta \right)}{2^n\sin \left(\theta \right)}$

Hence, the value of $\cos \left(\theta \right)·\cos \left(2\theta \right)·\cos \left(4\theta \right)·...·\cos \left(2^{n-1}\theta \right)=\frac{\sin \left(2^{\mathrm{n}}\theta \right)}{2^{\mathrm{n}}\sin \left(\theta \right)}$