Definitely, you can approach it via @Incredible Mind's Method but I would like to share another method.

We can proceed by using Euler's Formula
\[e^{ix} = \cos x + i\sin x\]
\[\Rightarrow \cos x = \frac{e^{ix}+e^{-ix}}{2}\]

Thus, setting \(x= \frac{2k\pi}{n}\), we can compute the sum as follows:

\[\begin{eqnarray}
\sum_{k=1}^{n-1} (n-k)\cos \frac{2k\pi}{n} & = & \sum_{k=1}^{n-1} (n-k) \frac{e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}}{2} \\
& = & \frac{n}{2} \sum_{k=1}^{n-1} \left(e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) - \frac{1}{2} \sum_{k=1}^{n-1} k \left( e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) \\
& = & \frac{n}{2} \sum_{k=1}^{n-1} \left(e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) - \frac{1}{2} \sum_{k=1}^{n-1} k \left( e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) \\
\end{eqnarray}\]
Now that above series (the first sum) is a simple Geometric Expansion and the second one can be obtained by differentiating the first one.
(Hint : \(\sum_{r=1}^{n} x^r = \frac{x(x^n-1)}{x-1}\) and \(\sum_{r=1}^{n} rx^r = \frac{x+(nx-n-1)x^{n+1}}{(x-1)^2}\))

Finally, it evaluates to
\[\sum_{k=1}^{n-1} (n-k)\cos \frac{2k\pi}{n} = \frac{n}{2}\left(-2\right)-\frac{1}{2}\left(-n\right) = -\frac{n}{2}\]
(I urge the reader to verify the above result themselves)

\(\textbf{EDIT : }\) A general approach for finding the sum of series of type \(\sum_{k=1}^n kr^k\) (doesn't involves differentiation)

Let \[S_n = \sum_{k=1}^n kr^k = r + 2r^2 + 3r^3 + \ldots + nr^n\]

To distinguish the 2 approaches, you should explain that the second summation can be obtained by methods other than differentiation, especially since Omkar said he doesn't know differentiation.

This is a useful formula to be aware of, to sum scenarios like this.

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TopNewestDefinitely, you can approach it via @Incredible Mind's Method but I would like to share another method.

We can proceed by using Euler's Formula \[e^{ix} = \cos x + i\sin x\] \[\Rightarrow \cos x = \frac{e^{ix}+e^{-ix}}{2}\]

Thus, setting \(x= \frac{2k\pi}{n}\), we can compute the sum as follows:

\[\begin{eqnarray} \sum_{k=1}^{n-1} (n-k)\cos \frac{2k\pi}{n} & = & \sum_{k=1}^{n-1} (n-k) \frac{e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}}{2} \\ & = & \frac{n}{2} \sum_{k=1}^{n-1} \left(e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) - \frac{1}{2} \sum_{k=1}^{n-1} k \left( e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) \\ & = & \frac{n}{2} \sum_{k=1}^{n-1} \left(e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) - \frac{1}{2} \sum_{k=1}^{n-1} k \left( e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) \\ \end{eqnarray}\] Now that above series (the first sum) is a simple Geometric Expansion and the second one can be obtained by differentiating the first one. (Hint : \(\sum_{r=1}^{n} x^r = \frac{x(x^n-1)}{x-1}\) and \(\sum_{r=1}^{n} rx^r = \frac{x+(nx-n-1)x^{n+1}}{(x-1)^2}\))

Finally, it evaluates to \[\sum_{k=1}^{n-1} (n-k)\cos \frac{2k\pi}{n} = \frac{n}{2}\left(-2\right)-\frac{1}{2}\left(-n\right) = -\frac{n}{2}\] (I urge the reader to verify the above result themselves)

\(\textbf{EDIT : }\) A general approach for finding the sum of series of type \(\sum_{k=1}^n kr^k\) (doesn't involves differentiation)

Let \[S_n = \sum_{k=1}^n kr^k = r + 2r^2 + 3r^3 + \ldots + nr^n\]

Then observing closely

\[\begin{eqnarray} S_n & = & r + 2r^2 + 3r^3 + \ldots + nr^n \\ rS_n & = & \qquad r^2 + 2r^3 + \ldots + (n-1)r^n + nr^{n+1} \end{eqnarray}\]

Subtracting the second series from second one, we get

\[(1-r)S_n = (r + r^2 + r^3 + \ldots + r^n) - nr^{n+1} = \frac{r(1-r^n)}{1-r}-nr^{n+1}\] \[\Rightarrow S_n =\frac{r(1-r^n)}{(1-r)^2}-\frac{nr^{n+1}}{1-r}\]

I hope that helps \(:)\)

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To distinguish the 2 approaches, you should explain that the second summation can be obtained by methods other than differentiation, especially since Omkar said he doesn't know differentiation.

This is a useful formula to be aware of, to sum scenarios like this.

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ok sir, I'll edit my solution, right now. \(:)\)

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Oh okay. Thanks!

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split it .then 1st can be known by using complex numbers.2nd sum from differentiation of

S=sinx+sin2x+sin3x...sin(n-1)x.........use the sum formula .then diff. w.r.t x both sides.then sub x=2pi/n.

i cannot type because i dont know LATEX

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Oh okay, but I don't know about differentiation. Is there any other method? I think complex numbers come handy.

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But there too you need to use differentiation.

Anyways, I'll definitely think of some another approach which doesn't involves differentiation.

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