Review question

# Why is $\cos \theta + \cos (\theta + 2\pi/3) + \cos(\theta + 4\pi/3)=0$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9782

## Solution

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.

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.

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

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*}