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 $\blacksquare$

~Yiyj1

1963 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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