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?

$\left(1+\frac1{1^2}\right)\left(1+\frac1{3^2}\right)\left(1+\frac1{5^2}\right)\left(1+\frac1{7^2}\right)\dotsm= \cosh \frac\pi 2$. SMS Product Series item no. 2 gives this product but omits the first factor.
