Question

Proof the following:
\[
\sum_{r=1}^{n}\sin\left(\frac{2\pi r}{n}\right)=0 \\
\sum_{r=1}^{n}\cos\left(\frac{2\pi r}{n}\right)=0
\]

Answer 1

Try not use complex numbers if possible.

Answer 2

do u know integration?

Answer 3

\[u ~can ~substitute ~(r/n)~as ~x\\then ~\it~will ~be\\ \int\limits_{0}^{1}\sin(2 \pi)x~dx\]
now just simple integration :)

Answer 4

I know what integration is but I don't know how you transformed a finite sum to a integral.

Answer 5

The even case is easy to prove but the odd case is hard.
@ganeshie8 @ikram002p

Answer 6

it will be something like this :-
if you start from - then you well end up with +
to get a sum of zero , i have to go :'( cant continue