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Correct Answer:
Correct answer was given by Manjunatha M R.
Note. If you find any mistake, please write to us at dep.math.svc@gmail.com
Remark. Before moving to the solution, we would like to make a remark that this is just a particular case of what is called Gauss trigonometric sum. Interested reasders may visit the book entitled Introduction to Analytic Number Theory by T. M. Apostol for more details about the Gauss sum.
We first present below the solution by Manjunatha M. R. and then present the original solution by the posers.
Solution by Manjunatha.
Define \(S\) and \(X\) as follows:\(S = \omega+\omega^2+\omega^4\)
and\(X = \omega^3+\omega^5+\omega^6\),
where \(\omega = e^{\frac{2\pi i }{7}}\). We know that
\(\omega^7 -1 =0\),
as \(\omega\) is the sevnth root of unity. The abov equation implies
\((\omega-1)(1+\omega+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6)=0\).
Since \(\omega-1\neq0\), we must have
\((1+\omega+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6)=0\).
That is
\(X+S+1=0\).
Or equivalently
\(X = -S-1\).
Now, it is easy to see that
\(S^2 = w+w^2+w^4+2w^3+2w^5+2w^6\).
That is
\(S^2 = S+2X = S+2(-1-S)\).
Thus, we will have
\(S^2+S+2=0\).
Solving for \(S\), we would arrive at
\(S = \frac{-1+i\sqrt{7}}{2} \quad \text{or} \quad \frac{-1-i\sqrt{7}}{2}.\)
To choose the correct value of \(S\), observe that
\(Im(S) = Im(\omega+\omega^2+\omega^4) = \sin\left(\frac{2\pi}{7}\right)+\sin\left(\frac{4\pi}{7}\right)+\sin\left(\frac{8\pi}{7}\right).\)
Observe that \(\sin\left(\frac{8\pi}{7}\right) = -\sin\left(\frac{\pi}{7}\right)\). Also
\(\sin\left(\frac{\pi}{7}\right) >0,\,\, \sin\left(\frac{2\pi}{7}\right)>0, \,\, \text{and}\,\, \sin\left(\frac{4\pi}{7}\right)>0\).
Further
\(\sin\left(\frac{2\pi}{7}\right)\) >\(\sin\left(\frac{\pi}{7}\right)\).
Hence
\( \sin\left(\frac{2\pi}{7}\right)+\sin\left(\frac{4\pi}{7}\right)+\sin\left(\frac{8\pi}{7}\right) >0\).
Hence, we must have
\(S = \frac{-1+i\sqrt{7}}{2}\).
This implies
\(X = -1-\frac{-1+i\sqrt{7}}{2}\).
Therefore,
\(S-X = 1+2S = i\sqrt{7}\).
Notice that \(S-X = \omega+\omega^2-\omega^3+\omega^4-\omega^5-\omega^6\). \(\blacksquare\)
Original Solution.
Let the required sum be denotes by \(T\). It is easy to notice that
\(T = \cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)-\cos\left(\frac{6\pi}{7}\right)+\cos\left(\frac{8\pi}{7}\right)-\cos\left(\frac{10\pi}{7}\right)-\cos\left(\frac{12\pi}{7}\right)\)
\(+i\left(\sin\left(\frac{2\pi}{7}\right)+\sin\left(\frac{4\pi}{7}\right)-\sin\left(\frac{6\pi}{7}\right)+\sin\left(\frac{8\pi}{7}\right)-\sin\left(\frac{10\pi}{7}\right)-\sin\left(\frac{12\pi}{7}\right)\right)\)
It is left to the readers to think why the following sum is 0 (without the help of any programming software):
\(\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)-\cos\left(\frac{6\pi}{7}\right)+\cos\left(\frac{8\pi}{7}\right)-\cos\left(\frac{10\pi}{7}\right)-\cos\left(\frac{12\pi}{7}\right)\)
Hence
\(T = i\left(\sin\left(\frac{2\pi}{7}\right)+\sin\left(\frac{4\pi}{7}\right)-\sin\left(\frac{6\pi}{7}\right)+\sin\left(\frac{8\pi}{7}\right)-\sin\left(\frac{10\pi}{7}\right)-\sin\left(\frac{12\pi}{7}\right)\right)\)
Following are starightforward consequences of multiple angle formulae:
\(\sin\left(\frac{2\pi}{7}\right)-\sin\left(\frac{12\pi}{7}\right) = 2\cos(\pi)\sin\left(\frac{-10\pi}{7}\right) = 2\sin\left(\frac{10\pi}{7}\right)\),
\(\sin\left(\frac{4\pi}{7}\right)-\sin\left(\frac{10\pi}{7}\right) = 2\cos(\pi)\sin\left(\frac{-6\pi}{7}\right)=2\sin\left(\frac{6\pi}{7}\right)\),
and
\(-\sin\left(\frac{6\pi}{7}\right)+\sin\left(\frac{8\pi}{7}\right) = 2\cos(\pi)\sin\left(\frac{2\pi}{7}\right)=-2\sin\left(\frac{2\pi}{7}\right)\).
Thus, \(T\) reduces to
\( T = 2i\left(-\sin\left(\frac{2\pi}{7}\right)+\sin\left(\frac{6\pi}{7}\right)+\sin\left(\frac{10\pi}{7}\right)-\sin\left(\frac{14\pi}{7}\right)\right)\).
Again the repeated application of multiple agnle formulae would yield
\(T = 8i\sin\left(\frac{\pi}{7}\right)\sin\left(\frac{2\pi}{7}\right)\sin\left(\frac{3\pi}{7}\right)\).
We agin leave to readers to show that
\(\sin\left(\frac{\pi}{7}\right)\sin\left(\frac{2\pi}{7}\right)\sin\left(\frac{3\pi}{7}\right) = \frac{\sqrt{7}}{8}\).
Hence \(T = i\sqrt{7}\).
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