Difference between revisions of "1963 IMO Problems/Problem 5"

Revision as of 19:40, 8 July 2021

Problem

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

Solutions

Solution 1

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 2

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. $\blacksquare$

@@ Line 36: / Line 36: @@
 Clearly <math>b\neq -1</math>, so <math>8b^3-4b^2-4b+1=0</math>. This proves the result. <math>\blacksquare</math>
-===Solution 3===
-Let <math>\omega=\mathrm{cis}\left(\frac{\pi}{14}\right)</math>. Thus it suffices to show that <math>\omega+\omega^{-1}-\omega^2-\omega^{-2}+\omega^3+\omega^{-3}=1</math>. Now using the fact that <math>\omega^k=\omega^{14+k}</math> and <math>-\omega^2=\omega^9</math>, this is equivalent to
-<cmath>\omega+\omega^3+\omega^5+\omega^7+\omega^9+\omega^{11}+\omega^{13}-\omega^7</cmath>
-<cmath>\omega\left(\frac{\omega^{14}-1}{\omega^2-1}\right)-\omega^7</cmath>
-But since <math>\omega</math> is a <math>14</math>th root of unity, <math>\omega^{14}=1</math>. The answer is then <math>-\omega^{7}=1</math>, as desired.
-~yofro
 ==See Also==
 {{IMO box|year=1963|num-b=4|num-a=6}}

1963 IMO (Problems) • Resources
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