Calculus $ {b \over n} \sum_{k=1}^n \sin\left({kb \over n}\right) = \text{?} $ First of all I want to mention that this is about an assignment I was given at school. I don't need the straight answer as much as a few hints to get started.
To give you a context, at the moment we're studying series and how they behave when $n$ goes to infinity.
The question I'm stuck at is to verify the following equality :
$$b \in {\Bbb R}, b \gt 0 ,~ n \in {\Bbb N}- \{0\} : \\ {b \over n}\sum_{k=1}^n \sin\left({kb \over n}\right) = 2\sin\left({b \over 2}\right){b / 2n \over {\sin\left(b \over 2n\right)}} \sin\left((n+1){b \over 2n}\right)$$
I feel like I'm missing an important trigonometric equation. Any help will be greatly appreciated !  Thanks a lot :)
 A: Ok guys, after a lot of infructuous search and with the help of Jack's link, I think I found the solution, I post it here for future reference and if anyone want to proof read it.
We start with :
$$ S ={b \over n}\sum_{k=1}^n sin\left({kb \over n}\right) $$
We multiply each side by $ sin({b\over2n}) $ and for convenience put it directly inside the sum :
$$ S.sin({b\over2n}) = {b \over n}\sum_{k=1}^n \left(sin\left({kb \over n}\right).sin({b\over2n})\right)  $$
We use the following trigonometric identity :
$$ sin(a).sin(b) = {1\over2}\left(cos(a-b)-cos(a+b)\right)  $$ 
Which gives us :
$$ S.sin({b\over2n}) = {b \over n}\sum_{k=1}^n{1\over2}\left(cos({kb \over n}-{b\over2n})-cos({kb \over n}+{b\over2n})\right) $$
We simplify the $cos()$ expressions :
$$ S.sin({b\over2n}) = {b \over n}\sum_{k=1}^n{1\over2}\left(cos({b\over2n}.(2k-1))-cos({b\over2n}.(2k+1))\right) $$
Now we can see it's a telescopic function, ie : of the form $ \sum_{k=1}^n a_k - a_{k+1}$ with $a_k = {b\over2n}(2k-1) $
So we can write : 
$$ S.sin({b\over2n}) = {b \over 2n}.\left( cos(a_1)-cos(a_{k+1}) \right) $$
$$ S.sin({b\over2n}) = {b \over 2n}.\left( cos({b\over2n})-cos\left({b\over2n}(2n+1)\right) \right) $$
$$ S.sin({b\over2n}) = {b \over 2n}.\left( cos({b\over2n})-cos\left(b+{b\over2n}\right) \right) $$
In the parenthesis we notice a second trigonometric identity : $$ cos(a) - cos(b) = -2.sin({a+b\over2}).sin({a-b\over2})$$
We apply this identity :
$$ S.sin({b\over2n}) = {b\over2n}.\left( -2.sin\left({b+{b\over n}\over 2}\right).sin\left(-{b\over 2}\right) \right)$$
Rearranging it a little (don't forget: $sin(-x) = -sin(x)$) we get to our final result : 
$$S = {b \over n}\sum_{k=1}^n sin\left({kb \over n}\right) = 2.sin\left({b \over 2}\right).{{b \over 2n} \over {sin\left(b \over 2n\right)}}.sin\left((n+1){b \over 2n}\right)$$
That was one of the hardest calculus I had to do in my life but cracking it felt awesome, thanks to everyone that helped me =)
A: What I have got for your sum is $$\frac{1-\cos b}{2}\frac{\frac{b}{n}\sin\frac{b}{n}}{1-\cos\frac{b}{n} }+\frac{b}{2n}\sin b$$
But I have used $\enspace\displaystyle \Im \frac{e^{ib}-1}{1-e^{-i\frac{b}{n}} }$ which comes from $\enspace\displaystyle \sum\limits_{k=1}^n x^k=\frac{x^n-1}{1-1/x}$ with $x:=e^{i\frac{b}{n}} $ .
