This is a modified summation question taken from HCI. I think it forces the students to think out of the box, and we all know how scary trigonometry and summation is together.

Given $\sum_{r=1}^{n}cos[(2r-1){\theta}] = \frac{sin(2n\theta)}{2sin\theta}$, show that

$sin{\theta}+3sin{3\theta}+5sin{5\theta}+...+99sin{99\theta} = \frac{cos{\theta}sin{100\theta}-100sin{\theta}cos{100\theta}}{2sin^{2}\theta}$

Hint: try writing out what you’re given and contrast it with what you are attempting to show. 🙂