Sum of the series $\sinθ\sin2θ + \sin2θ\sin3θ + \sin3θ\sin4θ + \sin4θ\sin5θ + \cdots+\sin n\theta\sin(n+1)\theta$ terms The series is given:
$$\sum_{i=1}^n \sin (i\theta) \sin ((i+1)\theta)$$
 We have to find the sum to n terms of the given series.
 I could took out the $2\sin^2\cos$ terms common in the series. But what to do further, please guide me.
 A: Use
$$2 \sin{n \theta} \, \sin{(n+1) \theta} = \cos{\theta}-\cos{(2 n+1)\theta}$$
Then the sum is
$$\frac12 n \cos{\theta} - \frac12 \sum_{k=1}^n \cos{(2 n+1)\theta}$$
The sum may be done as follows:
$$\begin{align} \sum_{k=1}^n \cos{(2 n+1)\theta} &= \operatorname{Re}{\left [e^{i \theta}\sum_{k=1}^n e^{i 2 k \theta}\right ]} \\ &=\operatorname{Re}{\left [e^{i 3 \theta} \frac{1-e^{i 2 n \theta}}{1-e^{i 2 \theta}}\right ]}\\ &= \operatorname{Re}{\left [e^{i (n+2) \theta} \frac{\sin{n \theta}}{\sin{\theta}}\right ]}\\ &= \frac{\sin{n \theta}}{\sin{\theta}} \cos{(n+2) \theta} \end{align}$$
A: Hint1:
$$\sum_{i=1}^n \sin (i\theta) \sin ((i+1)\theta)
=  1/2\sum_{i=1}^n \left(\cos (\theta)-\cos ((2i+1)\theta)\right)$$
Hint2:
$$ 2\sin (\theta)\sum_{i=1}^n \cos ((2i+1)\theta) = \sum_{i=1}^n \left(\sin ((2i+2)\theta) - \sin (2i\theta)\right) $$
A: Hint: For the part $A = \displaystyle \sum_{i=1}^{n} \cos (2i+1)x$, use:
$\sin x\displaystyle \sum_{i=1}^n \cos (2i+1)x$, and expand each term you can see a telescope sum.
