cos A cos 2 A cos 4 A ldots cos 2^{n-1} A equ | vedclass

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

Question

(A) To evaluate the product $P = \cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$,multiply and divide by $2 \sin A$: $P = \frac{2 \sin A \cos A \cos 2

Vedclass · Accepted Answer

$\frac{\sin 2^n A}{2^n \sin A}$

Vedclass · Answer

(A) To evaluate the product $P = \cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$,multiply and divide by $2 \sin A$: $P = \frac{2 \sin A \cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A}{2 \sin A}$ Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get: $P = \frac{\sin 2A \cos 2A \cos 4 A \ldots \cos 2^{n-1} A}{2 \sin A}$ Multiply and divide by $2$ again: $P = \frac{2 \sin 2A \cos 2A \cos 4 A \ldots \cos 2^{n-1} A}{2^2 \sin A} = \frac{\sin 4A \cos 4A \ldots \cos 2^{n-1} A}{2^2 \sin A}$ Repeating this process $n$ times,we obtain: $P = \frac{\sin 2^n A}{2^n \sin A}$

$\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ equals

Similar Questions

In a $\triangle ABC$,the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is equal to:

$A$ group of atoms can function as a ligand only when

The ratio of momenta of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of $100\, V$ is

Which of the following forms a stable $+4$ oxidation state?

Vedclass Test Series

Exam Paper Generator

Online Exam Module