Trigonometry> Cos(2pi/2n+1) +cos(4pi/2n+1)+cos(6pi/2na1...

Cos(2pi/2n+1) +cos(4pi/2n+1)+cos(6pi/2na1).......+cos(2npi/2n+1)

Abhinav Acharya , 10 Years ago

Grade 11

2 Answers

Arun Kumar

Last Activity: 10 Years ago

Hello Student,

let

x=2pi/(2n+1)

cosx+cos2x+cos3x+cos4x.......cosn=sin((n+1)x/2)cos(nx/2)/sin(x/2)

replace x by its value

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

Approved

Sumit Majumdar

Last Activity: 10 Years ago

Dear student,

You may use the following result:

Let $S=\cos\alpha+\cos 2\alpha+\ldots+\cos n\alpha$

Now multiply both members with $2\sin\frac{\alpha}{2}$

$2\sin\frac{\alpha}{2}S=2\sin\frac{\alpha}{2}\cos \alpha+2\sin\frac{\alpha}{2}\cos 2\alpha+\ldots+2\sin\frac{\alpha}{2}\cos n\alpha$

Using the identity $2\sin a\cos b=\sin(a+b)-\sin(b-a)$ we have

$2\sin\frac{\alpha}{2}S=\sin\frac{3\alpha}{2}-\sin\frac{\alpha}{2}+\sin\frac{5\alpha}{2}-\sin\frac{3\alpha}{2}+\ldots+\sin\frac{(2n+1) \alpha}{2}-\sin\frac{(2n-1)\alpha}{2}$

$2\sin\frac{\alpha}{2}S=\sin\frac{(2n+1)\alpha}{2}-\sin\frac{\alpha}{2}$

$2\sin\frac{\alpha}{2}S=2\sin\frac{n\alpha}{2}\cos \frac{(n+1)\alpha}{2}$

Then $S=\dfrac{\sin\frac{n\alpha}{2}\cos\frac{(n+1)\alph a}{2}}{\sin\frac{\alpha}{2}}$

Substitute $\alpha$ with $\frac{2\pi}{2n+1}$

Then $S=\dfrac{\sin \frac{n\pi}{2n+1} \cos \frac{(n+1) \pi}{2n+1}}{\sin\frac{\pi}{2n+1}}=$

$=\frac{1}{2}\cdot\dfrac{\sin\pi-\sin\frac{\pi}{2n+1}}{\sin\frac{\pi}{2n+1}}=\frac{ 1}{2}\cdot\left(-\dfrac{\sin\frac{\pi}{2n+1}}{\sin\frac{\pi}{2n+1}} \right)=-\frac{1}{2}$