Thread: Summation of sine and cosine

1. Summation of sine and cosine

I have a problem solving the following exercise.

i have to prove that $\sum_{j=0}^{n-1} cos (2j \pi /{n}})^2$ = $\sum_{j=0}^{n-1} sin (2j \pi /{n}})^2$ = 1/2

I have to prove that each sum is equal to 1/2 !!

(n is odd and I know that both sums together are 1, but I cannot prove why each is 1/2)

Help is appreciated!

2. is i natural?

3. i'm sorry, i will edit it, it is not natural

4. It's a simple application of Euler's Formula - $\displaystyle r\,e^{i\theta} = r\cos{\theta} + i\,r\sin{\theta}$...

5. I have to prove that the sums are equal to 1/2, that's my only problem. I think I didn't explain it right, I'm sorry.

6. Originally Posted by sssitex
I have a problem solving the following exercise.

i have to prove that $\sum_{j=0}^{n-1} cos (2j \pi /{n}})^2$ = $\sum_{j=0}^{n-1} sin (2j \pi /{n}})^2$ = 1/2

I have to prove that each sum is equal to 1/2 !!

(n is odd and I know that both sums together are 1, but I cannot prove why each is 1/2)

Help is appreciated!
Shouldn't that be n/2 instead of 1/2?

Hint: Can you show the first part? I.e.,

$\sum_{j=0}^{n-1} cos (2j \pi /{n}})^2$ = $\sum_{j=0}^{n-1} sin (2j \pi /{n}})^2$

7. I solved it, thanks

8. The 'geometrical sum' of the n squared 'roots of unity' is...

$\displaystyle \sum_{k=0}^{n-1} e^{2\ j\ \frac{2\ k\ \pi}{n}} = \frac{1-e^{4\ j\ \pi}}{1-e^{4\ j\ \frac{\pi}{n}}} =0$ (1)

... and from (1) we derive that is...

$\displaystyle \sum_{k=0}^{n-1} (\cos^{2} \frac{2\ k\ \pi}{n} - \sin^{2} \frac{2\ k\ \pi}{n}) =0$ (2)

Bur is also...

$\displaystyle \sum_{k=0}^{n-1} (\cos^{2} \frac{2\ k\ \pi}{n} + \sin^{2} \frac{2\ k\ \pi}{n}) =n$ (3)

... and from (2) and (3) we derive that is...

$\displaystyle \sum_{k=0}^{n-1} \cos^{2} \frac{2\ k\ \pi}{n} = \sum_{k=0}^{n-1} \sin^{2} \frac{2\ k\ \pi}{n} = \frac{n}{2}$ (4)

Kind regards

$\chi$ $\sigma$