Solve $\sum_{r=1}^n (2r-1)\cos(2r-1)\theta $

Question

How should i solve : $$\sum_{r=1}^n (2r-1)\cos(2r-1)\theta $$

I can solve $\sum_{r=1}^n cos(2r-1)\theta $ by considering $\Re \sum_{r=1}^n z^{2r-1} $ and using summation of geometric series, but I can't seem to find a common geometric ratio when $ 2r-1 $ is involved in the summation.

Visually : $\sum_{r=1}^n z^{2r-1} = z +z^3+...+z^{2r-1}$ where the common ratio $ r= z^2 $ can easily be seen, but in the case of $\sum_{r=1}^n (2r-1)z^{2r-1} = z + 3z^3 + 5z^5 +...+ (2r-1)z^{2r-1}$, how should i solve this ? A hint would be appreciated.

Answer 1

Note that $$\sum_{r=1}^{n} (2r-1) \cos (2r-1) \theta=\sum_{r=1}^{n} \frac{\mathrm{d}}{\mathrm{d}\theta}\sin (2r-1) \theta= \frac{\mathrm{d}}{\mathrm{d}\theta}\sum_{r=1}^{n} \sin (2r-1) \theta$$ Now calculate $$\sum_{r=1}^{n} \sin (2r-1) \theta=\frac{ \sin^2(n\theta) }{ \sin(\theta) }$$ Through the formula for the sum of sin's.

Now just diffferentiate with regard to $\theta $.

Answer 2

Here's a hint: $$ {d \over dz} z^n = n z^{n-1}$$. What do you get if you differentiate the geometric series term by term?