What is the logic behind the trick used to prove the summation series of Sine of Angles in Arithmetic Progression? 
why does multiplying by $2sin(b/2)$ cancel out most of the terms,what is the intuition behind it?how does multiplying by it creates a chain of terms that get cancelled?
edit:1 I have attached an image about what I tried to simplify using $sin(d)$
 A: *

*In the given example, it is observed that the original series can be
converted into a telescoping series, which is a series whose
general term is expressible as the difference of two consecutive terms of some sequence.
This is achieved by multiplying҂ it by $$2\sin\frac b2,$$ then
applying the product-to-sum
identity
$$2\sin\theta\sin\phi=\cos(\theta-\phi)-\cos(\theta+\phi).$$ The
resulting  telescoping series is
$$\sum_{r=1}^n\left[\cos\left(a+\frac{(2r-3)b}2\right)-\cos\left(a+\frac{(2r-1)b}2\right)\right].$$


*Next, the method of differences
is used to cancel out all the terms, except for the first and last
ones, of this telescoping series.


*Finally, the sum-to-product identity $$\cos\theta-\cos\phi=-2\sin\frac{\theta+\phi}2\sin\frac{\theta-\phi}2$$ is applied to express the result in terms of $\sin$ again.

҂Answer to the OP's further question:
Multiplying the series by $(2\sin b)$ instead of $(2\sin\frac b2)$ will result in the series $$\sum_{r=1}^n\left[\cos\left(a+(r-2)b\right)-\cos\left(a+rb\right)\right],$$ which isn't telescoping (refer the italicised text above); using the method of differences on this series will leave four instead of two terms.
