Question

Hloo ❤️ Ans plz
.....Que .... evaluate

plz don't post irrelevant answer otherwise it will be reported ❤️ ​

Answer 1

Answer :

1 - 1/√2

Solution :

We know that ,

sin2A = 2•sinA•cosA

Thus ,

$\sf sin x = 2. sin \dfrac{x}{2}. cos{x}{2}$

$\sf sin x = 2^2 sin \dfrac{x}{2^2} . cos \dfrac{x}{2^2} . cos \dfrac{x}{2}$

$\sf sin x = 2^3 sin \dfrac{x}{2^3} . cos \dfrac{x}{2^3} . cos \dfrac{x}{2^2} . cos \dfrac{x}{2}$

$\sf sin x = 2^n . sin \dfrac{x}{2^n}. \bigg[ cos \dfrac{x}{2} . cos \dfrac{x}{2^2} . cos \dfrac{x}{2^3} ........ cos \dfrac{x}{2^n}\bigg]$

$\sf sin x = 2^n sin \dfrac{x}{2^n} \bigg[ \prod\limits_{k=1}^{n} . cos \dfrac{x}{2^k} \bigg$

$\sf \prod\limits_{k=1}^{n} cos \dfrac{x}{2^k} = \dfrac{sin x}{2^n sin \frac{x}{2^n}}$

$\sf \prod\limits_{k=1}^{\infty} cos \dfrac{x}{2^k} = \int\limits_{n \arrow \infty} . \dfrac{sin x}{2^n . sin \frac{x}{2^n}}$

$\sf \prod\limits_{k=1}^{\infty} cos \dfrac{x}{2^k} = \dfrac{sin x}{x} . \int\limits_{n \arrow \infty} . \dfrac{\dfrac{1}{sin \frac{x}{2^n}}{\frac{x}{2^n}}$

$\sf \prod\limits_{k=1}^{\infty} cos \dfrac{x}{2^k} = \dfrac{sin x}{x}.1$

$\sf \prod\limits_{k=1}^{\infty} cos \dfrac{x}{2^k} = \dfrac{sin x}{x}$

Thus ,

$\sf \int\limits^{\pi /4}_{0} \prod\limits_{k=1}^{\infty} cos \dfrac{x}{2^k} dx = \int\limits^{\pi /4}_{0} x. \dfrac{sin x}{x}$

$\sf \int\limits^{\pi /4}_{0} sin x$

$[-cos x]^{\pi /4}_{0}$

$-[cos x]^{\pi /4}_{0}$

$\sf - \bigg[cos \dfrac{\pi}{4} - cos 0\bigg]$

$\sf - \bigg[\dfrac{1}{\sqrt{2}} - 1\bigg]$

$\sf 1 - \dfrac{1}{\sqrt{2}}$

Hence ,

The required answer is (1 - 1/√2) .