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

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

Grade:11

2 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
7 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
Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
7 years ago
Dear student,
You may use the following result:
LetS=\cos\alpha+\cos 2\alpha+\ldots+\cos n\alpha

Now multiply both members with2\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 identity2\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}

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

Substitute\alphawith\frac{2\pi}{2n+1}

ThenS=\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}
Regards
Sumit

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