Show that 3 sinusoidal phasors sum to zero 
Prove that $\sin(\theta) + \sin(\theta+2\pi/3) + \sin(\theta+4\pi/3) = 0 $ for any angle $\theta$.

This came up in the context of electricity:
It is common in electrical power engineering to use three-phase circuits with sinusoidal currents out of phase with each other by 120 degrees. The benefit of this is that the currents sum to zero at the neutral point meaning less copper is required for the system. 

However, while I know that the three phasors always sum to zero, and can see visually, as in the figure that this is true at every point, how can this be shown more formally, or algebraically?
 A: In complex numbers
$$e^{i\omega t}+e^{i\omega t+\phi}+e^{i\omega t+2\phi}=e^{i\omega t}(1+e^{i\phi}+e^{i2\phi})$$ where $\phi$ is $\frac{2\pi}3$.
As you easily check, $e^{i\phi}$ is a root of unit so that the sum between parenthesis is null.
Geometrically, the three complex numbers form an equilateral triangle centered, rotating around the origin at constant speed.
A: Here are some hints.
Use the addition formula:
$$\sin(a\pm b) = \sin(a)\cos(b) \pm \cos(a)\sin(b)$$
You can also note that the phase $4 \pi/3$ is also $-2 \pi / 3$, so you can take advantage of the fact that $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$.
Can you take it from here?  (Solution hidden below)

  $$S(x) \\ = \sin(x) + \sin(x)\cos(2 \pi/3) + \cos(x)\sin(2 \pi/3) + \sin(x)\cos(-2 \pi/3) + \cos(x)\sin(-2 \pi/3) \\ = \sin(x) + \sin(x)\cos(2 \pi/3) + \cos(x)\sin(2 \pi/3) + \sin(x)\cos(2 \pi/3) - \cos(x)\sin(2 \pi/3) \\ = \sin(x) + (-1/2)\sin(x) + \cos(x)\sin(2 \pi/3) + (-1/2)\sin(x) - \cos(x)\sin(-2 \pi/3) \\ = \sin(x) - \sin(x) \\ = 0.$$

A: The complex solution is quite simple. See that the points on the unit circle of argument $\theta, \theta+\frac{2\pi}{3}, \theta+\frac{4\pi}{3}$ are evenly spaced along the circle, and thus their sum as complex numbers is $0$. Your sines are the imaginary part of these three points, which also add to $0$.
A: We use complex numbers. Remember that
$$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$
and
$$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}.$$
Then,
\begin{align}
I &= \sin(\theta)+\sin(\theta + 2\pi/3)+ \sin(\theta + 4\pi/3) \\
&= \frac{e^{i\theta}-e^{-i\theta}}{2i}+\frac{e^{i(\theta+2\pi/3)}-e^{-i(\theta+2\pi/3)}}{2i}+\frac{e^{i(\theta+4\pi/3)}-e^{-i(\theta+4\pi/3)}}{2i}.
\end{align}
As 
$$e^{2\pi  i/3}=e^{-4\pi  i/3} \qquad \text{and} \qquad e^{-2\pi  i/3}=e^{4\pi  i/3}$$
We have that
$$\frac{e^{i\theta}e^{4\pi/3}-e^{-i\theta}e^{-4\pi/3}}{2i}=\frac{e^{i\theta}e^{-2\pi/3}-e^{-i\theta}e^{2\pi/3}}{2i}$$
The last two terms of $I$ are now
$$\frac{1}{2i}\left(e^{i\theta}(e^{2\pi i/3}+e^{-2\pi i/3})-e^{-i\theta}(e^{2\pi i/3}+e^{-2\pi i/3})\right).$$
But $(e^{2\pi i/3}+e^{-2\pi i/3})=-1$, so
$$I=\frac{e^{i\theta}-e^{-i\theta}}{2i}+\frac{1}{2i}\left(e^{i\theta}(e^{2\pi i/3}+e^{-2\pi i/3})-e^{-i\theta}(e^{2\pi i/3}+e^{-2\pi i/3})\right)=\frac{e^{i\theta}-e^{-i\theta}}{2i}+\frac{1}{2i}\left(-e^{i\theta}+e^{-i\theta}\right)=0$$
A: In general, we have for all $n\in\mathbb N$ with $n\geq2$ and $x\in\mathbb R$:
$$\bbox[10px,#ffd]{\sum_{k=1}^n \sin\left(x+\frac{2k\pi}n\right)}=\Im\left(\exp(ix)\sum_{k=1}^n\exp(2\pi i/n)^k \right)\overset{\text{geometric series}}=\Im\left(\exp(ix)\exp(2\pi i/n)\frac{1-\overbrace{\exp(2\pi i/n)^n}^1}{1-\exp(2\pi i/n)}\right)=\Im(0)=0.$$
Here, $\Im(z)$ denotes the imaginary part of $z\in\mathbb C$.
In your case, $n=3$.
