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.

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TopNewestIf I recall correctly, the official solution is to consider the expression \[ \sin \frac{\pi}{7} \left( \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} \right), \] distributing the sine factor and using the product-to-sum identity \[ \sin \alpha \cos \beta = \frac{1}{2} \left( \sin (\alpha+\beta) + \sin (\alpha - \beta) \right). \]

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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?

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

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Since \(\cos\frac\pi 7 = -\cos\frac{6\pi}7\) and \(\cos\frac{3\pi}7 = -\cos\frac{4\pi}7\), it suffices to prove \[\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{6\pi}7=-\frac 12\]

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

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Nice application!

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Thanks, very nicely explained.

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

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

Rearranging, we get

\[0=-1+\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}-\cos\frac{4\pi}{7}+\cos\frac{5\pi}{7}-\cos\frac{6\pi}{7}\]

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

\[0=\mathrm{Re}(-1+u-u^2+u^3-u^4+u^5-u^6)\]

But, since \(u^7-1=(u-1)(u^6-u^5+u^4-u^3+u^2-u+1)=0\), we know either \(u=1\) or \(u^6-u^5+u^4-u^3+u^2-u+1=0\). Clearly \(u\neq1\) since \(\mathrm{Re}(u)=\sin\frac{\pi}{7}\neq0\), so

\[u^6-u^5+u^4-u^3+u^2-u+1=0\]

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

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\[\cos \frac{ 2\pi }{ 7 }+\cos \frac{ 4\pi }{ 7 }+\cos \frac{ 6\pi }{ 7 }\]\[\frac{ \sin \frac{ 3\pi }{ 7 }*\cos \frac{ 3\pi }{ 7 } }{ \sin \frac{ \pi }{ 7 } }\]\[\frac{ \sin \frac{ 6\pi }{ 7 } }{ 2\sin \frac{ \pi }{ 7 } }\]\[\frac{ 1 }{ 2 }\]

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

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ill try this one... :D

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

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Recall that \[ \cos z = \frac{e^{iz} + e^{-iz}}{2}. \] Also, observe that \[ -\cos \frac{2\pi}{7} = \cos \frac{5\pi}{7}. \] Now let \( \zeta_n = e^{2i\pi/n} \) be a primitive \( n^{\rm th} \) root of unity. Thus \[ \begin{align*} \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} &= \cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} \\ &= \frac{1}{2}(\zeta_{14} + \zeta_{14}^{-1} + \zeta_{14}^3 + \zeta_{14}^{-3} + \zeta_{14}^5 + \zeta_{14}^{-5} ) \\ &= \frac{1}{2}(\zeta_{14} + \zeta_{14}^{13} + \zeta_{14}^3 + \zeta_{14}^{11} + \zeta_{14}^5 + \zeta_{14}^{9} ) \\ &= \frac{1}{2}\left( -\zeta_{14}^7 + \sum_{k=0}^6 \zeta_{14}^{2k+1} \right) \\ &= \frac{1}{2}\left( 1 + \zeta_{14} \sum_{k=0}^6 \zeta_7^k \right) \\ &= \frac{1}{2}(1 + \zeta_{14} \cdot 0) = \frac{1}{2}. \end{align*} \] Note that we used the following facts: \[ \zeta_{2n}^{2k} = \zeta_n^k, \] and because \( \{ 1, \zeta_n, \zeta_n^2, \ldots, \zeta_n^{n-1} \} \) are the roots of \( z^n - 1 = 0 \), the sum of these roots is the negative of the coefficient of the degree \( n-1 \) term, which is zero for any positive integer \( n \ge 2 \).

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

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right

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