Why is $\cos \theta + \cos (\theta + 2\pi/3) + \cos(\theta + 4\pi/3)=0$?

Why is $\cos \theta + \cos (\theta + 2\pi/3) + \cos(\theta + 4\pi/3)=0$?

Question

1. Prove that $\cos(A+B)=\cos A\cos B - \sin A\sin B,$ where the angles $A$, $B$ and $A+B$ are all acute.

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1. By projecting the sides of an equilateral triangle onto a certain line, prove that $\cos \theta + \cos \left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta+\frac{4\pi}{3}\right)=0$ and find the value of the expression $\sin \theta + \sin \left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta+\frac{4\pi}{3}\right).$

Suggestion

There is a standard geometric proof of this formula using the unit circle. If you haven’t come across it before, you may like to have a look at the diagram hidden below.

A useful diagram

What lengths can you calculate, in terms of $\alpha$ and $\beta$?

Could we draw an equilateral triangle at an angle $\theta$ to the horizontal? What angles do we know? What horizontal lengths can we deduce?

To find the value of the $\sin$ sum, what vertical lengths can we deduce?

Solution

We’ll use $\alpha$ and $\beta$ to represent the angles, and capital letters to represent points on the diagram. So we are trying to prove that

$$\cos(\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.$$

The circle drawn has radius $1$ (it’s a unit circle). From triangle $OCD$, we see that $\quad OD = \cos(\alpha+\beta)$.

From triangle $OFC$, we know $\quad CF = \sin \beta \quad$ and $\quad OF = \cos \beta.$

From triangle $CFG$, $\sin \alpha = \frac{FG}{FC} \quad \Longrightarrow \quad FG = \sin \alpha \sin \beta.$

From triangle $OEF$, $\cos \alpha = \frac{OE}{OF} \quad \Longrightarrow \quad OE = \cos \alpha \cos \beta.$

As $FG$ is equal in length to $DE$, we have

$$\begin{align*} OE &= OD + DE \\ &= OD + FG. \\ \end{align*}$$

Therefore $\cos \alpha \cos \beta = \cos(\alpha+\beta) + \sin \alpha \sin \beta,$ so $\cos(\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta,$ as required.

The restriction to acute angles is actually unnecessary - this works for all angles.

Let’s start by drawing an equilateral triangle inside a unit circle, at an angle $\theta$ to the horizontal.

We can quickly identify (using parallel lines and angles in triangles) that $\angle BAD = \theta + \frac{2 \pi}{3} \qquad \text{and} \qquad \angle CBO \text{ (reflex)} = \theta + \frac{4 \pi}{3}.$

Thinking about the unit circle, we can see (by considering the diagrams below) that

If we take $\psi = \theta + \dfrac{2 \pi}{3}$, we see that

$$\cos \left(\theta + \dfrac{2\pi}{3} \right) = \overrightarrow{AM} \equiv \overrightarrow{A'B'},$$

and taking $\phi = \theta + \dfrac{4\pi}{3}$, we have

$$\cos \left(\theta + \dfrac{4\pi}{3} \right) = \overrightarrow{B'O}.$$

Therefore

$$\cos \theta + \cos \left(\theta + \frac{2\pi}{3} \right) + \cos \left(\theta + \frac{4\pi}{3} \right) = \overrightarrow{OA'} + \overrightarrow{A'B'} + \overrightarrow{B'O}.$$

(i.e. we move from $O$ to $A'$, return to $B'$ and then back to $O$).

Therefore the total displacement is zero, and so $\cos \theta + \cos \left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta+\frac{4\pi}{3}\right)=0.$

…and find the value of the expression $\sin \theta + \sin \left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta+\frac{4\pi}{3}\right).$

The sum for $\sin$ can similarly be found by projecting the sides of an equilateral triangle onto the $y$-axis.

We can think again about the unit circle and how we find the sine of different angles:

Let $\psi = \theta + \dfrac{2\pi}{3}$ and $\phi = \theta + \dfrac{4\pi}{3}$. From our diagram above, we get

$$\begin{align*} \sin \theta + \sin \left(\theta + \frac{2\pi}{3} \right) + &\sin \left(\theta + \frac{4\pi}{3} \right) \\ &= \overrightarrow{A'A} + \overrightarrow{MB} + \overrightarrow{BB'} \\ &= \overrightarrow{B'M} + \overrightarrow{MB} + \overrightarrow{BB'} \\ &= 0. \end{align*}$$

Suppose we are allowed to use any method for the last two parts. If we know about complex numbers, they can be tackled more quickly. Let $w = \cos\theta + i\sin\theta$. Then if $k = e^{2\pi/3}$, we have $w+kw+k^2w = \cos \theta + \cos \left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta+\frac{4\pi}{3}\right)+i(\sin \theta + \sin \left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta+\frac{4\pi}{3}\right)).$ But we know $k^3 = 1$, so $k^3 -1 = (k-1)(k^2+k+1) =0$. Since $k \neq 1$, we have $k^2+k+1 = 0$, and so $\cos \theta + \cos \left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta+\frac{4\pi}{3}\right)+i(\sin \theta + \sin \left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta+\frac{4\pi}{3}\right))=0,$ which gives us both our results.