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I'm wondering if there is an analogous equation describing the relation between the Gamma function and cosines, as Euler's Reduction Formula does for the Gamma function and sines:

$$\Gamma (z) \Gamma (1-z)= \frac{ \pi }{\sin( \pi z)} $$

After looking and not seeing anything in the descriptions and identities for the Gamma function, I tried deriving it myself, starting with the trigonometric identity:

$$\sin( \pi z)^2 + \cos( \pi z)^2 = 1$$

and solving for $\cos (\pi z)$, however, a few steps into the derivation and get lost here:

$$\cos( \pi z) = \pm \frac{ \sqrt{(\Gamma (z) \Gamma (1-z))^{2} - \pi^{2} } }{ \Gamma (z) \Gamma (1-z)}$$

I know that $\Gamma(1/2) = \sqrt{\pi}$, so $(\Gamma(1/2))^{4} = \pi^{2}$ but that may not be helpful. I'm not sure where to go from here, or if there even is such a relation using cosines. If there is, I'd like to know what it is, or how to derive it, or at the very least, what it's called. Does this exist? Thanks!

EDIT I think I wrote the last line of my derivation incorrectly, and should read $\Gamma(1-z)$ not $\Gamma(z-1)$, which I have now changed.

UPDATE My understanding from researching this elsewhere is that formula does exist, and specifically exists in terms of Gamma functions of integers. The one answer posted is in terms of 1/2 plus the argument, which is not the version I'm looking for.

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If you just replace $z$ with $\frac{1}{2}-z$ you get: $$\Gamma\left(\frac{1}{2}-z\right)\Gamma\left(\frac{1}{2}+z\right)=\frac{\pi}{\cos(\pi z)}.$$

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  • $\begingroup$ I had tried that separately using the identities for Gamma(1/2 +/- n) and multiplying them but just got (-4)^n/(4^n). Did I miss something? Otherwise, could I have continued with my original proof and find it that way? And by the way, does this have a name? $\endgroup$ – iwantmyphd Jun 15 '15 at 15:52
  • $\begingroup$ Is there a version that uses just z and not 1/2-z? $\endgroup$ – iwantmyphd Oct 2 '15 at 16:37

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