# 1963 IMO Problems/Problem 5

## Problem

Prove that $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$.

## Solution

Because the sum of the $x$-coordinates of the seventh roots of unity is $0$, we have $$\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\frac{8\pi}{7}} + \cos{\frac{10\pi}{7}} + \cos{\frac{12\pi}{7}} + \cos{\frac{14\pi}{7}} = 0$$ $$\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\frac{8\pi}{7}} + \cos{\frac{10\pi}{7}} + \cos{\frac{12\pi}{7}} = -1$$ $$\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\left(-\frac{6\pi}{7}\right)} + \cos{\left(-\frac{4\pi}{7}\right)} + \cos{\left(-\frac{2\pi}{7}\right)} = -1.$$

Now, we can apply $\cos{x} = \cos{\left(-x\right)}$ to obtain $$2\left(\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}}\right) = -1$$ $$\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} = -\frac{1}{2}.$$

Finally, since $\cos{x} = -\cos{(\pi-x)}$, $$\cos{\frac{2\pi}{7}} - \cos{\frac{3\pi}{7}} - \cos{\frac{\pi}{7}} = -\frac{1}{2}$$ $$\cos{\frac{\pi}{7}} - \cos{\frac{2\pi}{7}} + \cos{\frac{3\pi}{7}} = \frac{1}{2}\textrm{. }\square$$

~mathboy100

## Solution 2

Let $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=S$. We have $$S=\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\cos{\frac{\pi}{7}}+\cos{\frac{3\pi}{7}}+\cos{\frac{5\pi}{7}}$$

Then, by product-sum formulae, we have $$S * 2* \sin{\frac{\pi}{7}} = \sin{\frac{2\pi}{7}}+\sin{\frac{4\pi}{7}}-\sin{\frac{2\pi}{7}}+\sin{\frac{6\pi}{7}}-\sin{\frac{4\pi}{7}}=\sin{\frac{6\pi}{7}}=\sin{\frac{\pi}{7}}$$

Thus $S = 1/2$. $\blacksquare$

## Solution 3

Let $a=\sin{\frac{\pi}{7}}$ and $b=\cos{\frac{\pi}{7}}$. From the addition formulae, we have $$S=b-[b^2-a^2]+[ b(b^2-a^2)-a(2ab) ]=b-b^2+b^3+a^2(1-3b)$$

From the Trigonometric Identity, $a^2=1-b^2$, so $$S=b-b^2+b^3+(1-b^2)(1-3b)=4b^3-2b^2-2b+1$$

We must prove that $S=1/2$. It suffices to show that $8b^3-4b^2-4b+1=0$.

Now note that $\cos{\frac{4\pi}{7}}=-\cos{\frac{3\pi}{7}}$. We can find these in terms of $a$ and $b$: $$\cos{\frac{4\pi}{7}}=[b^2-a^2]^2-[2ab]^2=b^4-6a^2b^2+b^4=b^4-6(1-b^2)b^2+b^4=8b^4-8b^2+1$$ $$\cos{\frac{3\pi}{7}}=b[b^2-a^2]-a[2ab]=b^3-3a^2b=b^3-3(1-b^2)b=3b-4b^3$$

Therefore $8b^4-8b^2+1=-(3b-4b^3)\Rightarrow 8b^4+4b^3-8b^2-3b+1=0$. Note that this can be factored: $$8b^4+4b^3-8b^2-3b+1=(8b^3-4b^2-4b+1)(b+1)=0$$

Clearly $b\neq -1$, so $8b^3-4b^2-4b+1=0$. This proves the result. $\blacksquare$

## Solution 4

Let $\omega=\mathrm{cis}\left(\frac{\pi}{14}\right)$. Thus it suffices to show that $\omega+\omega^{-1}-\omega^2-\omega^{-2}+\omega^3+\omega^{-3}=1$. Now using the fact that $\omega^k=\omega^{14+k}$ and $-\omega^2=\omega^9$, this is equivalent to $$\omega+\omega^3+\omega^5+\omega^7+\omega^9+\omega^{11}+\omega^{13}-\omega^7$$ $$\omega\left(\frac{\omega^{14}-1}{\omega^2-1}\right)-\omega^7$$ But since $\omega$ is a $14$th root of unity, $\omega^{14}=1$. The answer is then $-\omega^{7}=1$, as desired.

~yofro

## Solution 5

We let $\omega = \mathrm{cis} \ \left(\dfrac{2\pi}{7} \right)$. We therefore have $w^i$, where $0 \leq i \leq 6$, are the $7^{\text{th}}$ roots of unity. Since $\sum_{i = 0}^6 \omega^i = 0$, then $\sum_{i = 1}^6 \omega^i = -1$, so $\sum_{i = 1}^3 \mathrm{Re}(\omega^i) = -\dfrac{1}{2}$. Therefore, because $\mathrm{cis} \ \alpha = \cos \alpha + i \sin \alpha, \mathrm{Re}(\mathrm{cis} \ \alpha) = \cos \alpha$, so $$\cos \left(\dfrac{2\pi}{7} \right) + \cos \left( \dfrac{4\pi}{7} \right) + \cos \left(\dfrac{6\pi}{7} \right) = -\dfrac{1}{2}$$ $$\implies -\cos \left(\dfrac{2\pi}{7} \right) - \cos \left(\dfrac{4\pi}{7} \right) - \cos \left(\dfrac{6\pi}{7} \right) = \dfrac{1}{2}$$

Since $\cos \alpha = - \cos(\pi - \alpha)$, we have $\cos \left(\dfrac{\pi}{7} \right) - \cos \left(\dfrac{2\pi}{7} \right) + \cos \left(\dfrac{3\pi}{7} \right) = \dfrac{1}{2}$ and we are done $\blacksquare$