Q:

The limit of $2\frac{\sin\left(\frac{M\theta_{k}}{2}\right)\cos\left(\frac{1}{2}(1+M)\theta_{k}\right)}{\sin\left(\frac{\theta_{k}}{2}\right)} $

I have
$$\lambda_{k,N}=2\dfrac{\sin\Big(\dfrac{M\theta_{k}}{2}\Big) \cos\Big(\dfrac{1}{2}(1+M)\theta_{k}\Big)}{\sin\Big(\dfrac{\theta_{k}}{2}\Big)} $$
where $\theta_{k} \in [0, 2\pi]$, I can then let  $\theta_{k}=\dfrac{2\pi k}{N}$ for $k=0...N$,
I would like to find the limit of $\lambda_{k,N}$ when $N \to +\infty$ knowing that $M=M(N)$ and $\dfrac{M}{N}=0$ when $N \to +\infty$.
it would have been more interesting if I can compute the limit of $\lambda_{k,N}$ for any $k$ but it's not computable for some values of $\theta_{k}$.
Is it possible to find a closed form expression for the limit ?

A:

Using the identity
\begin{equation}
\sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin{(A-B)}]
\end{equation}
we get
\begin{eqnarray}
\lambda_{kN}&=&\frac{\sin\left[\left(M+\frac{1}{2}\right)\theta_k\right]-\sin\left(\frac{\theta_k}{2}\right)}{2\sin\left(\frac{\theta_k}{2}\right)}\\
&=&\frac{\sin\left((2M+1)\frac{\pi k}{N}\right)-\sin\left(\frac{\pi k}{N}\right)}{2\sin\left(\frac{\pi k}{N}\right)}\\
&=&\frac{\sin\left((2M+1)\frac{\pi k}{N}\right)}{2\sin\left(\frac{\pi k}{N}\right)}-\frac{1}{2}
\end{eqnarray}
So for any fixed $k$
\begin{equation}
\lim_{N\to\infty}\lambda_{kN}=\lim_{n\to\infty}M(N)+\frac{1}{2}-\frac{1}{2}=\lim_{n\to\infty}M(N)
\end{equation}
provided the limit exists.

