# Prove sum of $\cos(\pi/11)+\cos(3\pi/11)+...+\cos(9\pi/11)=1/2$ using Euler's formula

Prove that $$\cos(\pi/11)+\cos(3\pi/11)+\cos(5\pi/11)+\cos(7\pi/11)+\cos(9\pi/11)=1/2$$ using Euler's formula.

Everything I tried has failed so far.

Here is one thing I tried, but obviously didn't work. $$\Re e \{e^{\frac{\pi}{11}i}(1+e^{\frac{2\pi}{11}i}+e^{\frac{4\pi}{11}i}+e^{\frac{6\pi}{11}i}+e^{\frac{8\pi}{11}i}) \}=\frac{1}{2}$$ $$\Re e \{e^{\frac{\pi}{11}i}(1+\sqrt[11]{e^{2\pi i}}+\sqrt[11]{e^{4\pi i}}+\sqrt[11]{e^{6\pi i}}+\sqrt[11]{e^{8\pi i}}) \}=\frac{1}{2}$$ $$\Re e \{e^{\frac{\pi}{11}i}(1+\sqrt[11]{1}+\sqrt[11]{1}+\sqrt[11]{1}+\sqrt[11]{1}) \}=\frac{1}{2}$$ $$\Re e \{5e^{\frac{\pi}{11}i} \}=\frac{1}{2}$$ $$5\cos(\frac{\pi}{11})=\frac{1}{2}$$ Which isn't true :D

• You've used that, for example, $e^{\frac{2 \pi i}{11}} = 1$, which learly isn't true. Jan 15, 2015 at 11:41
• @rmico, See math.stackexchange.com/questions/117114/… Jan 15, 2015 at 11:45
• Why is this tagged (power-series)? Jan 15, 2015 at 14:01

Put $$S = \cos(π/11)+\cos(3π/11)+\cos(5π/11)+\cos(7π/11)+\cos(9π/11)$$

Then $$S = \cos(-π/11)+\cos(-3π/11)+\cos(-5π/11)+\cos(-7π/11)+\cos(-9π/11)$$ (because cos is even)

and of course $$-1 = \cos (-11\pi/11)$$

Sum these all up and you get $$2S-1$$ as the sum of the real parts of the eleven eleventh-roots of unity, which is 0, and therefore $$S = 1/2$$

• Awesome, although I have to use Euler's formula! But thanks a lot! Jan 15, 2015 at 11:55

HINT:

First of all, $$e^x=1\iff x=2n\pi$$ where $n$ is some integer

$\sum_{r=0}^5\cos\frac{(2r+1)\pi}{11}$ =Re[$\sum_{r=0}^5\left(e^{\frac{\pi i}{11}}\right)^{2r+1}$]

Now $\sum_{r=0}^5\left(e^{\frac{\pi i}{11}}\right)^{2r+1}$ is a Geometric Series

• I think the sum should be from 0 to 4, but I still don't get the answer TT I'm using the formula for geometric series, but I don't get 1/2. Jan 15, 2015 at 12:32
• Okay now I finally got it. Thanks so much! Jan 15, 2015 at 13:01

Sum of cosines of $n$ angles in Arithmetic Progression ( starting at $\alpha$, common angle difference $\beta$ ) has a ready formula, with simple trig derivation/simplification:

$$\cos \alpha + \cos ( \alpha + \beta )+ \cos ( \alpha + 2\beta) + \cos ( \alpha + (n-1) \beta = \dfrac { \sin (n \beta/2)} { \sin (\beta/2) }\cdot \cos ( \alpha + {(n+1)\beta/2} ).$$

The same is likewise valid for sines, the quotient does not change.

$$\sin \alpha + \sin ( \alpha + \beta )+ \sin ( \alpha + 2\beta) + \sin ( \alpha + (n-1) \beta = \dfrac { \sin (n \beta/2)} { \sin (\beta/2) }\cdot \sin ( \alpha + {(n+1)\beta/2} ).$$

EDIT1:

Multiply second equation with $i$ and add to the first one. Use Euler's relation in a geometric series with constant ratio terms, common difference is in the exponent.

• Thank you, but I had to use Euler's formula to find the answer. Jan 15, 2015 at 13:01
• OK, here you can use Euler's formula again to verify, after using it to derive at first. Jan 15, 2015 at 13:17