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?
$\endgroup$ 32 Answers
$\begingroup$$\cosh(x)$ is usually defined defined as $\frac{e^{x} + e^{-x}}{2}$. If you haven't some different definition, then it is quite straightforward: $$\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)$$
$\endgroup$ $\begingroup$$\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.
$\endgroup$ 1
