Question

$\forall n\in N$,
$\cos \theta \cos 2\theta \cos 4\theta \cos 8\theta \dots \cos \left(2^{n-1}\right)\theta =$

• A. $\frac{\sin 2^n\theta }{2^n\sin \theta }$
• B. $\frac{\cos 2^{n-1}\theta }{2^{n-1}\cos \theta }$
• C. $\frac{\sin 2^{n-1}\theta }{2^n\sin \theta }$
• D. $\frac{\cos 2^n\theta }{2^n\cos \theta }$

Answer 1

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

We know that $2\sin \theta \cos \theta =\sin 2\theta$
$a=\cos \theta \cos 2\theta \cos 4\theta \cos 8\theta ........\cos \left(2^{n-1}\right)\theta$
$\Rightarrow a=\frac{2^{n-1}\left(2\cos \theta \sin \theta \right)\cos 2\theta \cos 4\theta \cos 8\theta ........\cos \left(2^{n-1}\right)\theta }{2^n\sin \theta }$
$\Rightarrow a=\frac{2^{n-2}\left(2\cos 2\theta \sin 2\theta \right)\cos 4\theta \cos 8\theta ........\cos \left(2^{n-1}\right)\theta }{2^n\sin \theta }$
$\Rightarrow a=\frac{2\sin \left(2^{n-1}\right)\theta \cos \left(2^{n-1}\right)\theta }{2^n\sin \theta }=\frac{\sin 2^n\theta }{2^n\sin \theta }$
Hence, option 'A' is correct.