Another approach for calculating the sum $$
\sum_{n=0}^{\infty}(-\cos \beta)^{n} \frac{\sqrt{\pi} \Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{2 \Gamma\left(\frac{n}{2}+1\right)}, \beta\in(o,\pi)
$$
Answer:
$$\frac{\pi-2\arcsin\cos\beta}{2\sin\beta}=\frac{\beta}{\sin\beta}$$
I have an indirect proof， but it is more complicated. I believe there is a more elegant way to do this.
My complicated answer:
$$\int_{0}^{\frac{\pi}{2}} \frac{\mathrm{d} \alpha}{1+\cos \alpha \cos \beta} =\int_{0}^{\frac{\pi}{2}} \sum_{n=0}^{\infty}(-\cos \alpha \cos \beta)^{n} \mathrm{~d} \alpha \\
  =\sum_{n=0}^{\infty}(-\cos \beta)^{n} \int_{0}^{\frac{\pi}{2}} \cos ^{n} \alpha \mathrm{d} \alpha \\
  =\sum_{n=0}^{\infty}(-\cos \beta)^{n} \frac{\sqrt{\pi} \Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{2 \Gamma\left(\frac{n}{2}+1\right)} \\
$$
On the other hand:
$$\int \frac{d x}{1+\varepsilon \cos x}=\frac{1}{\sqrt{1-\epsilon^{2}}} \operatorname{arctan}\left(\sqrt{\frac{1-\varepsilon}{1+\varepsilon}} \operatorname{tan} \frac{x}{2}\right)+C, \varepsilon\in(-1,1)$$
In this way, I can get the answer. But it is complicate. So I need help.
 A: Using Legendre's duplication formula, $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\pi^{1/2}\Gamma(2z)$, your sum
$$\frac{\pi}{2}+\frac{\sqrt{\pi}}{2}\sum_{n=1}^{\infty}(-\cos \beta)^{n} \frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{ \Gamma\left(\frac{n}{2}+1\right)}$$
can be rewritten
$$
\frac{\pi}{2}+\frac{\sqrt{\pi}}{2}\sum_{n=1}^{\infty}(-\cos \beta)^{n}\frac{2^{1-n}\pi^{1/2}\Gamma(n)}{\Gamma(n/2)}\frac{2}{n\Gamma(n/2)}
=\frac{\pi}{2}+2\pi\sum_{n=1}^{\infty}x^n\frac{(n-1)!}{n\Gamma^2(n/2)}.
$$
where $x=-1/2\cos\beta$. Splitting into $n$ even/odd and applying the duplication formula again gives,
$$\frac{\pi}{2}+\left(2\pi\sum_{n=1}^{\infty}x^{2n}\frac{(2n-1)!}{2n(n-1)!^2}\right)+\left(2x+\frac{1}{8}\sum_{n=1}^{\infty}(4x)^{2n+1}\frac{(2n)!}{(2n+1)}\frac{(n-1)!^2}{(2n-1)!^2}\right).$$
The coefficients of powers of $x$ in the left hand sum appear to be A00170 and those of the right, $1/n$ multiples of reciprocals of Apéry numbers A005430, so we have
$$\frac{\pi}{2}+2x+\pi\sum_{n=1}^{\infty}{2n-1\choose n-1}x^{2n}+\sum_{n=1}^{\infty}\frac{1}{(n+1){2n+2\choose n+1}}(4x)^{2n+1}.$$
Mathematica has no problem evaluating these sums to obtain
$$\frac{\pi -\cos ^{-1}(2 x)}{\sqrt{1-4 x^2}}=\frac{\beta}{\sin\beta},$$
for $\beta\in(0,\pi)$. Check out this Binomial Sums link to see sums very similar to those here.
This might not be as elegant as you're looking for (evaluating the sums might not be considered elegant) but I spent some time on it so thought I'd show it anyway.
