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Math Challenge III Problem 2.32
Math Challenge III Problem 2.32
For the $\sum_{k = 1}^{n - 1}{k\cos{\frac{2k\pi}{n}}}$ part, I was able to find a solution by pairing up terms and using trigonometric identities. However, for the $\sum_{k = 1}^{n - 1}{k\sin{\frac{2k\pi}{n}}}$ part, I do not know what to do. Are there any good strategies that I could try? Thank you.
Re: Math Challenge III Problem 2.32
If you find a method for the $\cos$ related problem, usually you can use the same method for the $\sin$ part, with some similar trig identities.
This chapter is about complex numbers. So the question should be solvable with complex number techniques. The question 2.15 is a similar problem, and you can try to follow the solution to that problem and use a similar method.
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