Sum identity involving sin

Question

How one can prove that $$\sum_{k=1}^n(-1)^k\sin(2k\theta)=\cos(n\pi/2+\theta+n\theta)\sec\theta\sin(n\pi/2+n\theta)?$$ It looks difficult as there is sum on the other side and product of trigonometric functions on the other side. The book Which Way did the Bicycle Go gave a hint of induction but it looks difficult.

Answer 1

$$\begin{align*} \sum_{k=1}^{n} (-1)^{k} \sin(2k\theta) &= {} \mathrm{Im}\Bigg[ \sum_{k=1}^{n} \big[ -\exp(2i\theta) \big]^{k} \Big) \Bigg] \\[2mm] &= \mathrm{Im}\Bigg[ \big( -\exp(2i\theta) \big)\frac{1 - \Big( - \exp(2i\theta) \Big)^{n}}{1 + \exp(2i\theta)} \Bigg] \quad \mathrm{if} \; 1 + \exp(2i\theta) \neq 0 \\[2mm] &= \mathrm{Im}\Bigg[ \big(-\exp(2i\theta) \big)\frac{1 - \exp\big( 2i n \theta + n\pi \big)}{1+\exp\big( 2i\theta\big)} \Bigg] \\[2mm] &= \mathrm{Im}\Bigg[ \big(-\exp(2i\theta) \big)\frac{\exp\Big( i n \theta + \frac{n\pi}{2} \Big)}{\exp(i\theta)} \times \frac{\exp\Big( -i n \theta - \frac{n\pi}{2} \Big) - \exp\Big( i n \theta + \frac{n\pi}{2} \Big)}{\exp(-i\pi\theta) + \exp(i\theta)} \;\; \Bigg] \\[2mm] &= \mathrm{Im}\Bigg[ -\exp\Big( i (n+1)\theta + \frac{n\pi}{2} \Big)\frac{-2i\sin\Big(n\theta + \frac{n\pi}{2} \Big)}{2\cos( \theta )} \Bigg] \\[2mm] &= \cos\Big( (n+1)\theta + \frac{n\pi}{2} \Big)\frac{\sin\Big(n\theta + \frac{n\pi}{2} \Big)}{\cos(\theta)}. &= \end{align*} $$