N nietzsche Dec 2016 62 10 - Apr 6, 2017 #1 Is there any identity for $\tan(\pi z)$ in terms of closed analytical formulas removing the factor $\pi$ from the argument? I only know that \begin{eqnarray} \sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)} \end{eqnarray} so could work as well with a formula for $\cos(\pi z)$ in the same way?

Is there any identity for $\tan(\pi z)$ in terms of closed analytical formulas removing the factor $\pi$ from the argument? I only know that \begin{eqnarray} \sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(1-z)} \end{eqnarray} so could work as well with a formula for $\cos(\pi z)$ in the same way?

romsek Math Team Sep 2015 2,751 1,531 USA Apr 6, 2017 #2 $\cos(\pi z) = \sin\left(\pi \left(z + \dfrac 1 2 \right)\right)$ $\cos(\pi z ) = \dfrac{\pi}{\Gamma\left(z+\frac 1 2 \right)\Gamma\left(\frac 1 2 - z\right)}$ $\tan(\pi z) = \dfrac{\Gamma\left(z+\frac 1 2 \right)\Gamma\left(\frac 1 2 - z\right)}{\Gamma(z)\Gamma(1-z)}$ Reactions: 2 people

$\cos(\pi z) = \sin\left(\pi \left(z + \dfrac 1 2 \right)\right)$ $\cos(\pi z ) = \dfrac{\pi}{\Gamma\left(z+\frac 1 2 \right)\Gamma\left(\frac 1 2 - z\right)}$ $\tan(\pi z) = \dfrac{\Gamma\left(z+\frac 1 2 \right)\Gamma\left(\frac 1 2 - z\right)}{\Gamma(z)\Gamma(1-z)}$