The Buffling Cosines

Can someone please help me answering this problem. This problem came from the IMO (i think its 1962). Here it goes:

Prove that \(\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} = \frac{1}{2}\)

I have done everything i know with this, even using the sum to product identities, still it doesn't work. Please help. thanks.

#OlympiadMath #Proofs #MathProblem #Math

Note by Jaydee Lucero
7 years, 9 months ago

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12 votes

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Comments

If I recall correctly, the official solution is to consider the expression sin⁡π7(cos⁡π7−cos⁡2π7+cos⁡3π7), \sin \frac{\pi}{7} \left( \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} \right), distributing the sine factor and using the product-to-sum identity sin⁡αcos⁡β=12(sin⁡(α+β)+sin⁡(α−β)). \sin \alpha \cos \beta = \frac{1}{2} \left( \sin (\alpha+\beta) + \sin (\alpha - \beta) \right).

hero p. - 7 years, 9 months ago

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wow!!! :D now i know how will I apply the product to sum identity. But, how should I remove the sine factor thereafter?

Jaydee Lucero - 7 years, 9 months ago

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After you compute and simplify the expression, you will find it equals 12sin⁡π7 \frac{1}{2} \sin \frac{\pi}{7} , from which the original identity immediately follows.

hero p. - 7 years, 9 months ago

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@Hero P. – and now I got it!!! haha :D thanks , men...

Jaydee Lucero - 7 years, 9 months ago

Since cos⁡π7=−cos⁡6π7\cos\frac\pi 7 = -\cos\frac{6\pi}7 and cos⁡3π7=−cos⁡4π7\cos\frac{3\pi}7 = -\cos\frac{4\pi}7, it suffices to prove cos⁡2π7+cos⁡4π7+cos⁡6π7=−12\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{6\pi}7=-\frac 12

But, 0=1+cos⁡2π7+cos⁡4π7+⋯+cos⁡12π7=1+2(cos⁡2π7+cos⁡4π7+cos⁡6π7)0=1+\cos\frac{2\pi}7+\cos\frac{4\pi}7+\dots+\cos\frac{12\pi}7=1+2\left(\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{6\pi}7\right) because 1,cos⁡2π7,cos⁡4π7,…,cos⁡12π71,\cos\frac{2\pi}7,\cos\frac{4\pi}7,\dots,\cos\frac{12\pi}7 are the real parts of the 7th roots of unity. The result follows.

Brice Huang - 7 years, 9 months ago

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Thanks, very nicely explained.

Lokesh Sharma - 7 years, 9 months ago

Nice application!

Calvin Lin Staff - 7 years, 9 months ago

Recall that cos⁡(π−k)=−cos⁡k\cos(\pi-k)=-\cos k. This means

cos⁡π7−cos⁡2π7+cos⁡3π7−(cos⁡6π7−cos⁡5π7+cos⁡4π7)=1\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}-\left(\cos\frac{6\pi}{7}-\cos\frac{5\pi}{7}+\cos\frac{4\pi}{7}\right)=1

Rearranging, we get

0=−1+cos⁡π7−cos⁡2π7+cos⁡3π7−cos⁡4π7+cos⁡5π7−cos⁡6π70=-1+\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}-\cos\frac{4\pi}{7}+\cos\frac{5\pi}{7}-\cos\frac{6\pi}{7}

Letting u=eiπ7u=e^\frac{i\pi}{7}, this is

0=Re(−1+u−u2+u3−u4+u5−u6)0=\mathrm{Re}(-1+u-u^2+u^3-u^4+u^5-u^6)

But, since u7−1=(u−1)(u6−u5+u4−u3+u2−u+1)=0u^7-1=(u-1)(u^6-u^5+u^4-u^3+u^2-u+1)=0, we know either u=1u=1 or u6−u5+u4−u3+u2−u+1=0u^6-u^5+u^4-u^3+u^2-u+1=0. Clearly u≠1u\neq1 since Re(u)=sin⁡π7≠0\mathrm{Re}(u)=\sin\frac{\pi}{7}\neq0, so

u6−u5+u4−u3+u2−u+1=0u^6-u^5+u^4-u^3+u^2-u+1=0

which completes the proof since Re(0)=0\mathrm{Re}(0)=0 as desired.

Cody Johnson - 7 years, 8 months ago

Use the 7th root of unity...I'm sure you will get the answer..

Rushi Rokad - 7 years, 9 months ago

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How? The cube root of unity would involve i i but there is no ii in the equation to be proved.

Lokesh Sharma - 7 years, 9 months ago

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Recall that cos⁡z=eiz+e−iz2. \cos z = \frac{e^{iz} + e^{-iz}}{2}. Also, observe that −cos⁡2π7=cos⁡5π7. -\cos \frac{2\pi}{7} = \cos \frac{5\pi}{7}. Now let ζn=e2iπ/n \zeta_n = e^{2i\pi/n} be a primitive nth n^{\rm th} root of unity. Thus cos⁡π7−cos⁡2π7+cos⁡3π7=cos⁡π7+cos⁡3π7+cos⁡5π7=12(ζ14+ζ14−1+ζ143+ζ14−3+ζ145+ζ14−5)=12(ζ14+ζ1413+ζ143+ζ1411+ζ145+ζ149)=12(−ζ147+∑k=06ζ142k+1)=12(1+ζ14∑k=06ζ7k)=12(1+ζ14⋅0)=12. \begin{aligned} \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} \\ &= \frac{1}{2}(\zeta_{14} + \zeta_{14}^{-1} + \zeta_{14}^3 + \zeta_{14}^{-3} + \zeta_{14}^5 + \zeta_{14}^{-5} ) \\ &= \frac{1}{2}(\zeta_{14} + \zeta_{14}^{13} + \zeta_{14}^3 + \zeta_{14}^{11} + \zeta_{14}^5 + \zeta_{14}^{9} ) \\ &= \frac{1}{2}\left( -\zeta_{14}^7 + \sum_{k=0}^6 \zeta_{14}^{2k+1} \right) \\ &= \frac{1}{2}\left( 1 + \zeta_{14} \sum_{k=0}^6 \zeta_7^k \right) \\ &= \frac{1}{2}(1 + \zeta_{14} \cdot 0) = \frac{1}{2}. \end{aligned} Note that we used the following facts: ζ2n2k=ζnk, \zeta_{2n}^{2k} = \zeta_n^k, and because {1,ζn,ζn2,…,ζnn−1} \{ 1, \zeta_n, \zeta_n^2, \ldots, \zeta_n^{n-1} \} are the roots of zn−1=0 z^n - 1 = 0 , the sum of these roots is the negative of the coefficient of the degree n−1 n-1 term, which is zero for any positive integer n≥2 n \ge 2 .

I think that you should be able to tell which solution method I prefer.

hero p. - 7 years, 9 months ago

ill try this one... :D

Jaydee Lucero - 7 years, 9 months ago

cos⁡2π7+cos⁡4π7+cos⁡6π7\cos \frac{ 2\pi }{ 7 }+\cos \frac{ 4\pi }{ 7 }+\cos \frac{ 6\pi }{ 7 }sin⁡3π7∗cos⁡3π7sin⁡π7\frac{ \sin \frac{ 3\pi }{ 7 }*\cos \frac{ 3\pi }{ 7 } }{ \sin \frac{ \pi }{ 7 } }sin⁡6π72sin⁡π7\frac{ \sin \frac{ 6\pi }{ 7 } }{ 2\sin \frac{ \pi }{ 7 } }12\frac{ 1 }{ 2 }

Cody Martin - 7 years, 8 months ago

right

Sonu Kumar Tiwari - 7 years, 8 months ago
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