Prove that $\Gamma(\frac{1}{2}-x)\Gamma(\frac{1}{2}+x)=\frac{\pi}{\cos\pi x}$

Question

I have to prove this equality $$\Gamma(\frac{1}{2}-x)\Gamma(\frac{1}{2}+x)=\frac{\pi}{\cos\pi x}$$
I assume, since both Gamma's have $\frac12$ as part of their argument, I'll have to use the fact that $\Gamma(\frac12)=\sqrt\pi$ I just don't know how. I also thought of using $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x}$.

Answer 1

Putting an answer here for the sake of closing the question.

Since $$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x},$$ it follows that $$\Gamma(\tfrac{1}{2}-x)\Gamma(\tfrac{1}{2}+x)=\Gamma(\tfrac{1}{2}-x)\Gamma(1-(\tfrac{1}{2}-x))=\frac{\pi}{\sin\pi(\tfrac{1}{2}-x)}=\frac{\pi}{\cos\pi x},$$ which is the desired solution.