This note is to provide a proof for

\[\sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12\]

**Proof:**

\(\begin{align} S & = \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) \\ & = \Re \left \{ \sum_{k=0}^{n-1} e^{\frac {2k+1}{2n+1}\pi i} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \sum_{k=0}^{n-1} e^{\frac {2k\pi i}{2n+1}} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \left(\frac {1-e^{\frac {2n\pi i}{2n+1}}}{1-e^{\frac {2\pi i}{2n+1}}}\right) \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}}-e^{\pi i}}{1-e^{\frac {2\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}} + 1}{\left(1+e^{\frac {\pi i}{2n+1}} \right) \left(1-e^{\frac {\pi i}{2n+1}} \right)} \right \} \\ & = \Re \left \{ \frac 1{1-e^{\frac {\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac 1{1-\cos \frac {\pi}{2n+1} - i\sin \frac {\pi}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi}{2n+1} + i\sin \frac {\pi}{2n+1}}{\left(1-\cos \frac {\pi}{2n+1}\right)^2 + \sin^2 \frac {\pi}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi}{2n+1} + i\sin \frac {\pi}{2n+1}}{2-2\cos \frac {\pi}{2n+1}} \right \} \\ & = \frac 12 \ \square \end{align} \)

Since \(\cos (\pi - x) = - \cos x\), we have: \(\displaystyle \sum_{k=0}^{n-1} \cos \left(\pi - \frac {2k+1}{2n+1}\pi \right) = \sum_{k=0}^{n-1} \cos \left(\frac {2n-2k}{2n+1}\pi \right) = \sum_{k=0}^{n-1} \cos \left(\frac {2k}{2n+1}\pi \right) = - \frac 12\).

## Comments

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TopNewestGiven what you've written up, can you provide a one-line proof to demonstrate that \[\sum_{k=0}^{n-1} \cos \left( \frac {2k}{2n+1}\pi \right) = -\frac 12? \] – Pi Han Goh · 3 weeks, 3 days ago

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– Chew-Seong Cheong · 3 weeks, 2 days ago

Thanks, now I see it.Log in to reply

good. A minor typo in the next-to-last three lines: delete the \(i\) from the sine and cosine args that resulted from a LATEX copy-and-paste error. – Wesley Zumino · 3 weeks, 3 days ago

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– Chew-Seong Cheong · 3 weeks, 3 days ago

Thanks a lot.Log in to reply