Question

For all $n\in N,$ $\cos \theta \cos 2\theta \cos 4\theta ......\cos 2^{n-1}\theta$ equals to

• A. $\frac{sin{2}^n\theta }{2^{n}sin\theta }$
• B. $\frac{sin{2}^n\theta }{sin\theta }$
• C. $\frac{cos{2}^n\theta }{2^{n}cos2\theta }$
• D. $\frac{cos{2}^n\theta }{2^{n}sin\theta }$

Answer 1

The correct option is A $\frac{sin{2}^n\theta }{2^{n}sin\theta }$

$\cos \theta \cos 2\theta \cos 4\theta ......\cos 2^{n-1}\theta$

$=\frac{\sin \theta \cos \theta \cos 2\theta \cos 4\theta ......\cos 2^{n-1}\theta }{\sin \theta }$

$=\frac{\left(2\sin \theta \cos \theta \right)\cos 2\theta \cos 4\theta ......\cos 2^{n-1}\theta }{2\sin \theta }$

$=\frac{\left(\sin 2\theta \right)\cos 2\theta \cos 4\theta ......\cos 2^{n-1}\theta }{2\sin \theta },[\because 2\sin x\cos x=\sin 2x]$

$=\frac{\left(2\sin 2\theta \cos 2\theta \right)\cos 4\theta \cos 8\theta ......\cos 2^{n-1}\theta }{2^2\sin \theta }$

$=\frac{\left(\sin 4\theta \right)\cos 8\theta \cos 16\theta ......\cos 2^{n-1}\theta }{2^2\sin \theta }$

similarly doing $n$ times

$=\frac{2\sin \left(2^{n-1}\theta \right)\cos \left(2^{n-1}\theta \right)}{2^n\sin \theta }=\frac{\sin \left(2^n\theta \right)}{2^n\sin \theta }$