If cos x + cos 2x + ldots + cos nx = frac{A(x | vedclass

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

Question

(B) Given, $\cos x + \cos 2x + \ldots + \cos nx = \frac{A(x)}{2 \sin(x/2)}$.Using the sum formula for cosine series, $\sum_{k=1}^n \cos(kx) = \frac{\s

Vedclass · Accepted Answer

$\frac{-4n}{2n+1}$

Vedclass · Answer

(B) Given, $\cos x + \cos 2x + \ldots + \cos nx = \frac{A(x)}{2 \sin(x/2)}$.Using the sum formula for cosine series, $\sum_{k=1}^n \cos(kx) = \frac{\sin(nx/2) \cos((n+1)x/2)}{\sin(x/2)}$.Thus, $A(x) = 2 \sin(nx/2) \cos((n+1)x/2)$.Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$A(x) = \sin((n + n/2 + 1/2)x) + \sin((n/2 - (n+1)/2)x) = \sin(\frac{2n+1}{2}x) - \sin(x/2)$.Now, $\int_0^\pi A(x) dx = \int_0^\pi (\sin(\frac{2n+1}{2}x) - \sin(x/2)) dx$.$= [-\frac{2}{2n+1} \cos(\frac{2n+1}{2}x) + 2 \cos(x/2)]_0^\pi$.$= (-\frac{2}{2n+1} \cos(\frac{2n+1}{2}\pi) + 2 \cos(\pi/2)) - (-\frac{2}{2n+1} \cos(0) + 2 \cos(0))$.Since $\cos(\frac{2n+1}{2}\pi) = 0$ and $\cos(\pi/2) = 0$:$= 0 - (-\frac{2}{2n+1} + 2) = \frac{2}{2n+1} - 2 = \frac{2 - 4n - 2}{2n+1} = \frac{-4n}{2n+1}$.

If $\cos x + \cos 2x + \ldots + \cos nx = \frac{A(x)}{2 \sin(x/2)}$, then $\int_0^\pi A(x) dx =$

Similar Questions

Evaluate the following integral: $\int_{1}^{2} \frac{x \, dx}{(x+1)(x+2)}$

If $\int \limits_{\frac{1}{3}}^3 |\log_e x| dx = \frac{m}{n} \log_e \left(\frac{n^2}{e}\right)$,where $m$ and $n$ are coprime natural numbers,then $m^2 + n^2 - 5$ is equal to $............$.

If $f(x) = |x - 1|$,then $\int_0^2 {f(x)dx}$ is

$\int_0^\pi \frac{1}{1+\sin x} dx$ is equal to

If $f(x)=|x-1|+|x-2|+|x-3|, \forall x \in[1,4]$,then $\int_1^4 f(x) dx=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module