Prove the relation of $\cosh(\pi/2)$ and $e$ Prove that:
$$\cosh\left(\frac{\pi}{2}\right)=\frac{1}{2}e^{-{\pi/2}}(1+e^\pi)$$
What I have tried.
$$\cosh\left(\frac{\pi}{2}\right)=\cos\left(i\frac{\pi}{2}\right)$$
$$=Re\{e^{i.i\frac{\pi}{2}}\}$$
$$=Re\{e^{-\frac{\pi}{2}}\}$$
Why is $e^{-\frac{\pi}{2}}$ not answer any why is $$\frac{e^{-\frac{\pi}{2}}+e^{\frac{\pi}{2}}}{2}$$
a correct solution. Did I miss something somewhere?

$\cosh(x)$ is usually defined defined as $\frac{e^{x} + e^{-x}}{2}$. If you haven't some different definition, then it is quite straightforward:
$$\cosh\left(\frac{\pi}{2}\right)=\frac{e^{\frac{\pi}{2}} + e^{-\frac{\pi}{2}}}{2} = \frac{1}{2}e^{-\frac{\pi}{2}}(1 + e^x)$$
