The value of cos frac{pi}{2^2} cdot cos frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We use the identity $\cos A \cdot \cos 2A \cdot \cos 4A \cdot \dots \cdot \cos 2^{n-1}A = \frac{\sin(2^n A)}{2^n \sin A}$.Let $A = \frac{\pi}{2^{1

Vedclass · Accepted Answer

$\frac{1}{512}$

Vedclass · Answer

(A) We use the identity $\cos A \cdot \cos 2A \cdot \cos 4A \cdot \dots \cdot \cos 2^{n-1}A = \frac{\sin(2^n A)}{2^n \sin A}$.Let $A = \frac{\pi}{2^{10}}$. The given expression is $P = \cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \dots \cdot \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$.Note that $\frac{\pi}{2^2} = 2^8 \cdot \frac{\pi}{2^{10}}$,$\frac{\pi}{2^3} = 2^7 \cdot \frac{\pi}{2^{10}}$,and so on.This is equivalent to $\cos(2^8 A) \cdot \cos(2^7 A) \cdot \dots \cdot \cos(A) \cdot \sin A$.Rearranging the product: $(\cos A \cdot \cos 2A \cdot \dots \cdot \cos 2^8 A) \cdot \sin A$.Using the formula with $n=9$: $\frac{\sin(2^9 A)}{2^9 \sin A} \cdot \sin A = \frac{\sin(2^9 \cdot \frac{\pi}{2^{10}})}{2^9} = \frac{\sin(\frac{\pi}{2})}{512} = \frac{1}{512}$.

The value of $\cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \dots \cdot \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$ is

Similar Questions

If $A+B=45^{\circ}$ and $(\cot A-1)(\cot B-1)=4 K$,then $K=$

The value of $\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+5 \tan ^{2} 30^{\circ}+\frac{3}{2}\sin ^{2} 90^{\circ}-3 \cos ^{2} 90^{\circ}$ is

The value of $x,$ which satisfies the equation $2 \operatorname{cosec}^{2} 30^{\circ} + x \sin^{2} 60^{\circ} - \frac{3}{4} \tan^{2} 30^{\circ} = 10$ is

If $\cot 3A = x,$ then the value of $x$ is

An observer on the top of a cliff,$200\, m$ above the sea level,observes the angle of depression of two ships at anchor to be $45^{\circ}$ and $30^{\circ}$ respectively. Find the distance between the ships,if the line joining them stretches to the base of the cliff. (in $m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module