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Prove that

$\sin(\theta) + \sin(\theta+2\pi/3) + \sin(\theta+4\pi/3) = 0 $

I was able to solve it using trigonometric identities but I want a method which uses complex numbers.

Any help would be appreciated.

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    See http://math.stackexchange.com/questions/2135044/prove-that-cos2-20-cos2-40-cos2-80-sin2-20-sin2-40-si/2135388#2135388 – 2017-02-13

2 Answers 2

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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 $0$.

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    could you please explain a little more, I am a little new to complex numbers. – 2017-02-13
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Remember that

$$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$

and

$$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}.$$

Then,

$$I=\sin(\theta)+\sin(\theta + 2\pi/3)+ \sin(\theta + 4\pi/3)=\frac{e^{i\theta}-e^{-i\theta}}{2i}+\frac{e^{i(\theta+2\pi/3)}-e^{-i(\theta+2\pi/3)}}{2i}+\frac{e^{i(\theta+4\pi/3)}-e^{-i(\theta+4\pi/3)}}{2i}.$$

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

$$I=\frac{e^{i\theta}-e^{-i\theta}}{2i}+\frac{1}{2i}\left(e^{i\theta}(e^{2\pi i/3}+e^{-2\pi i/3})-e^{-i\theta}(e^{2\pi i/3}+e^{-2\pi i/3})\right)=\frac{e^{i\theta}-e^{-i\theta}}{2i}+\frac{1}{2i}\left(-e^{i\theta}+e^{-i\theta}\right)=0$$