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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 :)

Lee David Chung Lin
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Furrane
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    you can verify it by using Lagrange's trigonometric identities : https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Lagrange.27s_trigonometric_identities – Sahil Kumar Sep 30 '16 at 16:51
  • Have a look at http://math.stackexchange.com/questions/1931509/how-to-find-area-under-sines-without-calculus – Jack D'Aurizio Sep 30 '16 at 17:09
  • Thanks guys for your help, Jack using your link I was able to solve it, this is awesome =) – Furrane Sep 30 '16 at 19:49

2 Answers2

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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 =)

Furrane
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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}} $ .

user90369
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