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{\sin

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

If $2f(x) - 3f\left( \frac{1}{x} \right) = x$,then $\int_1^2 f(x) \, dx$ is equal to

Let $f(x) = \max \{x+|x|, x-[x]\}$,where $[x]$ denotes the greatest integer $\leq x$. Then,the value of $\int_{-3}^{3} f(x) dx$ is

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

$\int_{0}^{4}|x-2| d x=$

If $\int \limits_{-0.15}^{0.15} |100 x^2 - 1| dx = \frac{k}{3000}$,then $k$ is equal to $..........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module