Simplify: $\sin \frac{2\pi}{n} +\sin \frac{4\pi}{n} +\ldots +\sin \frac{2\pi(n-1)}{n}$.

Question

Can you help me solve this problem?

Simplify: $\sin \dfrac{2\pi}{n} +\sin \dfrac{4\pi}{n} +\ldots +\sin \dfrac{2\pi(n-1)}{n}$.

Answer 1

The sum of all $n$th roots of unity (for $n > 1$) is zero. See here for the proof. Its imaginary part is also zero. That is,

$$\sum_{k = 0}^{n - 1} \sin \frac{2\pi k}{n} = 0$$

Now simply subtract $\sin 0 = 0 $ from both sides to get

$$\sum_{k = 1}^{n - 1} \sin \frac{2\pi k}{n} = 0$$

Answer 2

Take the terms in opposite pairs, and note the change of sign,

$$\sin \dfrac{2k\pi}{n}+\sin \dfrac{2\pi(n-k)}{n}=\sin \dfrac{2k\pi}{n} +\sin(2\pi-\dfrac{2k\pi}{n})=\sin \dfrac{2k\pi}{n} -\sin\dfrac{2k\pi}{n}=0.$$ In case that $n$ is even, the central term remains, but $$\sin \dfrac{2n\pi}{2n}=0.$$

Answer 3

$\sin k\alpha=\frac{\cos (k-1)\alpha-\cos (k+1)\alpha}{2\sin\alpha}$ by using compound angle formula given $\sin\alpha \neq0$. Then take summation the numerator will cancel. In your case $\alpha=\frac{2\pi}{n}, k=1\cdots n-1$. So you also need to discuss the case $n=1,2$ separately.

Answer 4

There is a formula if the angles of sine are in A.P.

$sinA +sin(A+D)....+sin(A+(n-1)D)=\frac{sin(nD/2)*sin(A+\frac{(n-1)D}{2})}{sin(D/2)}$.

Use it to get $sin2\pi$ in the numerator to get your answer 0.

Also, a complex approach might do as well. Sum of $n^{th}$ roots of unity is 0. So it's imaginary part is 0 as well.