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.

No vote yet

12 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 \[ \sin \alpha \cos \beta = \frac{1}{2} \left( \sin (\alpha+\beta) + \sin (\alpha - \beta) \right). \]

Log in to reply

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

Log in to reply

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.

Log in to reply

Log in to reply

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.

Log in to reply

Thanks, very nicely explained.

Log in to reply

Nice application!

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

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.

Log in to reply

ill try this one... :D

Log in to reply

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

Log in to reply

right

Log in to reply