The sum of this 'geometric series'...
(1)
... allows You to write...
(2)
(3)
Kind regards
Find simple formulas for: 1 + cos(x) + cos(2x) + ... + cos(nx)
and: sin(x) + sin(2x) + ... + sin(nx).
Remember that: 1 + a + a^2 + ... + a^n = (a^(n+1) - 1) / (a - 1)
and: (cosx + i*sinx)^n = cos(nx) + i*sin(nx).
So far, I showed that cos(nx) = (cosx + i*sinx)^n - i*sin(nx) and plugged that in for the first sum to get:
1 + cos(x) + cos(2x) + ... + cos(nx)
= 1 + (cosx + i*sinx) - i*sin(x) + (cosx + i*sinx)^2 - i*sin(2x) + ... + (cosx + i*sinx)^n - i*sin(nx)
= ((cosx + i*sinx)^(n+1) - 1)/(cosx + i*sinx - 1) - (i*sin(x) + i*sin(2x) + ... + i*sin(nx))
I'm not sure where to proceed from here to simplify it any further.