Summation of the trigonometric series

Question

If $\alpha$ is the exterior angle of a regular polygon of $n$ sides , prove that $$\sum_{r=0}^{n-1}\sin (\theta+ \alpha r ) = 0\; .$$

Answer 1

So, we need to prove that for every $\theta\in\mathbb{R}$: $$\sum_{r=0}^{n-1}\sin(\theta+\alpha r)=0$$ Actually, we will prove something stronger; that the some apllies for $\cos$ as well: $$\sum_{r=0}^{n-1}\cos(\theta+\alpha r)=0$$ Now, at first, note, as Alex Vong pointed out, that $\alpha=\frac{2\pi}{n}$. Now, the idea of the proof is quite geometric. Let us consider the following vectors on the Cartesian plane: $$\vec{v_r}=(\cos(\theta+\alpha r),\sin(\theta+\alpha r)),\ r=0,1,\dots,n-1$$ These form, as one can see, a regular polygon inscribed in the unitary circle. For our ease, we will use complex numbers to make use of some elementary properties of complex multiplication. So, using Euler's formula: $$e^{ix}=\cos x+i\sin x$$ we can describe each of our vectors $\vec{v_r}$ as a complex number. More precisely: $$\vec{v_r}\mapsto e^{i(\theta+\alpha r)}$$ Now, let us sum these vector: $$\begin{align*} S&=\sum_{r=0}^{n-1}e^{i(\theta+\alpha r)}=\sum_{r=0}^{n-1}e^{i\theta}e^{i\alpha r}=\\ &=e^{i\theta}\sum_{r=0}^{n-1}e^{i\alpha r}=e^{i\theta}\sum_{r=0}^{n-1}\left(e^{i\alpha}\right)^r\tag{Geometric sum}\\ &=e^{i\theta}\frac{1-\left(e^{i\alpha}\right)^n}{1-e^{i\alpha}}=\\ &=e^{i\theta}\frac{1-e^{2\pi i}}{1-e^{i\alpha}}=\\ &=e^{i\theta}\frac{1-1}{1-e^{i\alpha}}=\\ &=0 \end{align*}$$ Now, note that: $$S=\sum_{r=0}^{n-1}e^{i\theta+\alpha r}=\sum_{r=0}^{n-1}\cos(\theta+\alpha r)+i\sin(\theta+\alpha r)=\sum_{r=0}^{n-1}\sin(\theta+\alpha r)+i\sum_{r=0}^{n-1}\cos(\theta+\alpha r)$$ But, a complex number is zero exactly when both it imaginary and real part are zero, so: $$S=0\Leftrightarrow\left\{\begin{array}{l} \displaystyle\sum\limits_{r=0}^{n-1}\sin(\theta+\alpha r)=0\\ \displaystyle\sum\limits_{r=0}^{n-1}\cos(\theta+\alpha r)=0 \end{array}\right.$$ So, the requested has been proved.