How does one show that $\sum_{k=1}^{n}\sin\left({\pi\over 2}\cdot{4k-1\over 2n+1}\right)={1\over 2\sin\left({\pi\over 2}\cdot{1\over 2n+1}\right)}?$ Consider 

$$\sum_{k=1}^{n}\sin\left({\pi\over 2}\cdot{4k-1\over 2n+1}\right)=S\tag1$$
  How does one show that $$S={1\over 2\sin\left({\pi\over 2}\cdot{1\over 2n+1}\right)}?$$

An attempt:
Let $$A={\pi\over 2(2n+1)}$$
$$\sin(4Ak-A)=\sin(4Ak)\cos(A)-\sin(A)\cos(4Ak)\tag2$$
$$\cos(A)\sum_{k=1}^{n}\sin(4Ak)-\sin(A)\sum_{k=1}^{n}\cos(4Ak)\tag3$$
$$\sin(4Ak)=4\sin(Ak)\cos(Ak)-8\sin^3(Ak)\cos(Ak)\tag4$$
$$\cos(4Ak)=8\cos^4(Ak)-8\cos^2(Ak)+1\tag5$$
Substituting $(4)$ and $(5)$ into $(3)$ is going to be quite messy. So how else can we prove $(1)$?
 A: Using Lagrange Identities: 
$$ \sum_{n=1}^{N}\sin\left(\theta n\right)=+\frac{\cot\left(\theta/2\right)}{2}-\frac{\cos\left(\theta N+\theta/2\right)}{2\sin\left(\theta/2\right)} \quad\&\quad \sum_{n=1}^{N}\cos\left(\theta n\right)=-\frac{1}{2}+\frac{\sin\left(\theta N+\theta/2\right)}{2\sin\left(\theta/2\right)}$$ 
One could simplify $(3)$ through a different path: 
$$ 
\begin{align} 
\color{red}{S} &= \sum_{k=1}^{n}\sin\left(\frac{\pi}{2}\,\frac{4k-1}{2n+1}\right) = \sum_{k=1}^{n}\sin\left(4A k-A\right) \\[3mm] 
&= \cos\left(A\right)\sum_{k=1}^{n}\sin\left(4A k\right)-\sin\left(A\right)\sum_{k=1}^{n}\cos\left(4A k\right) \\[3mm] 
&= \cos\left(A\right)\left[\frac{\cot\left(2A\right)}{2}-\frac{\cos\left(4A n+2A\right)}{2\sin\left(2A\right)}\right]+\sin\left(A\right)\left[\frac{1}{2}-\frac{\sin\left(4A n+2A\right)}{2\sin\left(2A\right)}\right] \\[3mm] 
&= \frac{\cos\left(A\right)\cos\left(2A\right)+\sin\left(A\right)\sin\left(2A\right)}{2\sin\left(2A\right)}-\frac{\cos\left(A\right)\cos\left(4A n+2A\right)+\sin\left(A\right)\sin\left(4A n+2A\right)}{2\sin\left(2A\right)} \\[3mm] 
&= \frac{\cos\left(2A-A\right)}{2\sin\left(2A\right)}-\frac{\cos\left(4A n+2A-A\right)}{2\sin\left(2A\right)} \qquad\qquad\qquad\left\{\color{blue}{\small 4A n+2A = 2(2n+1)A = \pi}\right\} \\[3mm] 
&= \frac{\cos\left(A\right)}{2\sin\left(2A\right)}-\frac{\cos\left(\pi-A\right)}{2\sin\left(2A\right)} = \frac{\cos\left(A\right)}{2\sin\left(2A\right)}+\frac{\cos\left(A\right)}{2\sin\left(2A\right)} \qquad\left\{\color{blue}{\small \cos(\pi-A)=-\cos(A)}\right\} \\[3mm] 
&= \frac{\cos\left(A\right)}{\sin\left(2A\right)} = \frac{\cos\left(A\right)}{2\sin\left(A\right)\cos\left(A\right)} = \frac{1}{2\sin\left(A\right)} = \color{red}{\frac{1}{2\sin\left(\frac{\pi/2}{2n+1}\right)}} 
\end{align} 
$$
A: Following the idea suggested to pass the problem to the complex:
$$\sum_{k=1}^{n}\sin\left({\pi\over 2}\cdot{4k-1\over 2n+1}\right)=\Im\left(\sum_{k=1}^{n}\exp\left(i{\pi\over 2}\cdot{4k-1\over 2n+1}\right)\right)$$
Now, let us manipulate the exponentials directly.
$$\exp\left(i{\pi\over 2}\cdot{4k-1\over 2n+1}\right)=\exp\left(i{\pi\over 2}\cdot{1\over 2n+1}\right)\exp\left(i{\pi\over 2}\cdot{4k-2\over 2n+1}\right)$$
$$\sum_{k=1}^{n}\exp\left(i{\pi\over 2}\cdot{4k-1\over 2n+1}\right)=\exp\left(i{\pi\over 2}\cdot{1\over 2n+1}\right)\sum_{k=1}^{n}\exp\left(i{\pi\over 2}\cdot{4k-2\over 2n+1}\right)$$
Define $\theta=i{\pi\over 2}\cdot{1\over 2n+1}$ and $z=e^{2\theta}$. We can sum the geometric series.
$$\exp\left(i{\pi\over 2}\cdot{1\over 2n+1}\right)\sum_{k=1}^{n}\exp\left(i{\pi\over 2}\cdot{4k-2\over 2n+1}\right)=e^{\theta}\sum_{k=1}^ne^{2\theta(2k-1)}=$$
$$=e^\theta\sum_{k=1}^nz^{(2k-1)}=e^\theta\frac{1-z^{(2n+1)}}{1-z^2}=$$
$$=e^\theta\frac{1-e^{i\pi(2n+1)/(2n+1)}}{1-e^{4\theta}}=e^\theta\frac{1-e^{i\pi}}{1-e^{4\theta}}=\frac{2e^\theta}{1-e^{4\theta}}=$$
$$=\frac{2e^\theta}{e^{2\theta}(e^{-2\theta}-e^{2\theta})}=\frac{2e^{-\theta}}{e^{-2\theta}-e^{2\theta}}=$$
$$=\frac{2(\cos\theta-i\sin\theta)}{-2i\sin(2\theta)}=i\frac{\cos\theta}{\sin{2\theta}}+\frac{\sin\theta}{\sin{2\theta}}=$$
$$=i\frac{\cos\theta}{2\sin\theta\cos\theta}+\frac{\sin\theta}{2\sin\theta\cos\theta}=\frac{i}{2\sin\theta}+\frac{1}{2\cos\theta}$$
Now, take the imaginary part and reverse the changes, and we are done.
$$\sum_{k=1}^{n}\sin\left({\pi\over 2}\cdot{4k-1\over 2n+1}\right)=\Im\left(\frac{i}{2\sin\theta}+\frac{1}{2\cos\theta}\right)$$
$$\sum_{k=1}^{n}\sin\left({\pi\over 2}\cdot{4k-1\over 2n+1}\right)=\frac{1}{2\sin\left({\pi\over 2}\cdot{1\over 2n+1}\right)}$$
