How to simplify the expression and get a closed form?
$$\frac{\sin\alpha\cos^2\beta}{\sin\beta}+\frac{\sin\beta\cos^2\alpha}{\sin\alpha}$$ where $\alpha := \frac12(m+1)\theta$ and $\beta:=\frac12(m-1)\theta$; and where $m$ is an integer.
I wanna get rid of the denominators or make it a constant without $m$, but don't know how.
On
If we have $z=\exp(i\theta/2)$ then this is \begin{align} &\frac{(z^{m+1}-z^{-(m+1)})(z^{m-1}+z^{-(m-1)})^2}{4(z^{m-1}-z^{-(m-1)})}+\frac{(z^{m-1}-z^{-(m-1)})(z^{m+1}+z^{-(m+1)})^2}{4(z^{m+1}-z^{-(m+1)})}\\ &=\frac{(z^{m+1}-z^{-(m+1)})^2(z^{m-1}+z^{-(m-1)})^2+ (z^{m-1}-z^{-(m-1)})^2(z^{m+1}+z^{-(m+1)})^2} {4(z^{m-1}-z^{-(m-1)})(z^{m+1}-z^{-(m+1)})}\\ &=\frac{(z^{2m}-z^2+z^{-2}-z^{-2m})^2+ (z^{2m}+z^2-z^{-2}-z^{-2m})^2} {4(z^{2m}-z^2-z^{-2}+z^{-2m})}\\ &=\frac{(z^{2m}-z^{-2m})^2+(z^2-z^{-2})^2} {2(z^{2m}-z^2-z^{-2}+z^{-2m})}\\ &=-\frac{\sin^2 m\theta+\sin^2\theta}{\cos m\theta-\cos \theta}. \end{align} I haven't got rid of the denominator, but it is an an alternative (simpler?) form. One could cancel a $\sin\theta$ but that seems to complicate things...
On
\begin{align}\frac{\sin\frac{m+1}{2}\theta\cos^2\frac{m-1}{2}\theta}{\sin\frac{m-1}{2}\theta}+\frac{\sin\frac{m-1}{2}\theta\cos^2\frac{m+1}{2}\theta}{\sin\frac{m+1}{2}\theta}&=\frac{\left( \sin \frac{m+1}2\theta \cos \frac{m-1}2\theta\right)^2+\left( \sin \frac{m-1}2\theta\cos\frac{m+1}2\theta\right)^2}{\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}\\ &=\frac{\left( \sin \frac{m+1}2\theta \cos \frac{m-1}2\theta\right)^2+\left( \sin \frac{m-1}2\theta\cos\frac{m+1}2\theta\right)^2}{\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}\\ &=\frac{\left( \sin m\theta +\sin \theta\right)^2+\left( \sin m\theta-\sin\theta\right)^2}{4\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}\\ &=\frac{\sin^2 m\theta + \sin^2 \theta}{{2\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}} \\ &=\frac{\sin^2 m\theta + \sin^2 \theta}{\cos \theta- \cos m\theta} \end{align}
From this desmos link, the vertical asymptotes is a function of $m$. Hence $m$ is a term in the denominator.
The vertical asymptotes (which due to dision by $0$ in the denominator) appear where $\sin \left( \frac{m-1}2 \theta\right)=0$ or $\sin \left( \frac{m+1}2 \theta\right)=0$, that is when
$$\frac{m \pm 1}2 \theta = k\pi, k \in \mathbb{Z}$$
$$\theta = \frac{2k\pi}{m \pm 1} , k \in \mathbb{Z}$$
which is a function of $m$. The denominator is dependent on $m$.
On
Let $2a = (m+1) \theta$ and $2b = (m-1) \theta$ to obtain $$f = \frac{\sin(a) \, \cos^{2}(b)}{\sin(b)} + \frac{\sin(b) \, \cos^{2}(a)}{\sin(a)}.$$ Now, consider the following: \begin{align} f &= \frac{1}{\sin(a) \, \sin(b)} \, \left(\sin^{2}(a) \, \cos^{2}(b) + \sin^{2}(b) \, \cos^{2}(a) \right) \end{align} Using $2 \sin^{2} x = 1 - \cos(2 x)$ and $2 \cos^{2} x = 1 + \cos(2 x)$, then \begin{align} f &= \frac{1}{\sin(a) \, \sin(b)} \, \left(\sin^{2}(a) \, \cos^{2}(b) + \sin^{2}(b) \, \cos^{2}(a) \right) \\ &= \frac{1}{4 \sin(a) \sin(b)} \, ( (1-\cos(2a))(1+\cos(2b)) + (1-\cos(2b))(1+\cos(2a)) ) \\ &= \frac{1 - \cos(2a) \cos(2b)}{2 \, \sin(a) \sin(b)} \\ &= \frac{2 - \cos2(a-b) - \cos2(a+b)}{4 \, \sin(a) \sin(b)} \\ &= \frac{\sin^{2}(a-b) + \sin^{2}(a+b)}{2 \, \sin(a) \sin(b)} \end{align}
This can be seen in the form $$f = \frac{\sin^{2}(m \theta) + \sin^{2}\theta}{\cos\theta - \cos(m \theta)}.$$
It becomes evident from $\sin(a) \, \sin(b)$ that there are problems (asymptotes) at $$\theta \in \left\{ \frac{2 n \pi}{m \pm 1} \right\} \hspace{5mm} n \geq 0. $$
I'm not sure if you can get rid of the denominator, but you can try to use these identities.