Question

find the value of 2^n cosx/2.cosx/4.cosx/8......cosx/2^n

Answer 1

We have to simplify

$f(x)=2^n \cos(x/2).\cos(x/4).\cos(x/8)......\cos(x/2^n)$.

Use the formula $2\sin \alpha \cos \alpha =\sin (2\alpha )$

Now,

$f(x) \sin(x/2)=2^n \cos(x/2)\sin(x/2).\cos(x/4).\cos(x/8)......\cos(x/2^n)\\ f(x) \sin(x/2^n)=2^n \cos(x/2)\sin(x/2).\cos(x/4).\cos(x/8)......[\cos(x/2^n) \sin(x/2^n)]\\ f(x) \sin(x/2^n)=2^{n-1} \cos(x/2)\sin(x/2).\cos(x/4).\cos(x/8)......[2\cos(x/2^n) \sin(x/2^n)]\\ f(x) \sin(x/2^n)=2^{n-1} \cos(x/2)\sin(x/2).\cos(x/4).\cos(x/8)......\cos(x/2^{n-1}) \sin(x/2^{n-1})$

$f(x) \sin(x/2^n)=2^{n-2} [\cos(x/2)\sin(x/2)].\cos(x/4).\cos(x/8)......[2\cos(x/2^{n-1}) \sin(x/2^{n-1})] \\ f(x) \sin(x/2^n)=2^{n-2} [\cos(x/2)\sin(x/2)].\cos(x/4).\cos(x/8)......\cos(x/2^{n-2}) \sin(x/2^{n-2})$

Proceeding in this way,

$f(x) \sin(x/2^n)=2\cos(x/2)\sin(x/2)\\ f(x) \sin(x/2^n)=\sin x\\ f(x) =\frac{\sin x}{\sin(x/2^n)}$