Cosine Summation Identity

**ralphjason007** · May 9th 2011, 05:26 PM

prove that: cos(2pi/2n+1)+cos(4pi/2n+1)+cos(6pi/2n+1)+...+cos(2npi/2n+1)=-1/2

**TheCoffeeMachine** · May 9th 2011, 07:41 PM

Originally Posted by ralphjason007

prove that: cos(2pi/2n+1)+cos(4pi/2n+1)+cos(6pi/2n+1)+...+cos(2npi/2n+1)=-1/2

Consider the equation $f(z) = z^{2n}+z^{2n-1}+\cdots+z+1 = 0.$
By the formula for geometric series, $f(z) = \frac{z^{2n+1}-1}{z-1}$ for $z \ne 1.$
Now, if $z^{2n+1}-1 = 0$ then either $z = 1$ or $e^{\pm \frac{2k\pi}{2n+1}i}$ where $1 \le k \le n.$
Thus the roots of $f(z)$ are $z = e^{\pm \frac{2k\pi}{2n+1}i}$ , where $1\le k \le n.$
The sum of these roots is $2\sum_{1 \le k \le n}\cos\left(\frac{2k\pi}{2n+1}\right).$
But by Vieta's relations the sum of the roots of $f(z)$ is $-1.$
Thus $2\sum_{1 \le k \le n}\cos\left(\frac{2k\pi}{2n+1}\right)= -1$ and hence:

$\sum_{1 \le k \le n}\cos\left(\frac{2k\pi}{2n+1}\right)= -\frac{1}{2}.$

**red_dog** · May 17th 2011, 07:52 AM

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

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

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

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