# How do I prove that $\sin(π/2+iy)=1/2(e^{y}+e^{−y})=\cosh y$?

How do I prove that $\sin(π/2+iy)=1/2(e^{y}+e^{−y})=\cosh y$?

It might surprise you to find that the addition formula for $\sin$ works for complex arguments as well... use that, along with $\sin\,ix=i\sinh\,x$ and $\cos\,ix=\cosh\,x$. –  Guess who it is. May 8 '12 at 2:17
The definition of $\cosh(y)$ is $$\cosh(y)=\frac{e^y+e^{-y}}{2}.$$ The definition of $\sin(z)$ (or a property, if you use some other definition) is $$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}.$$ Thus $$\sin(\tfrac{\pi}{2}+iy)=\frac{e^{i\left(\tfrac{\pi}{2}+iy\right)}-e^{-i\left(\tfrac{\pi}{2}+iy\right)}}{2i}=\frac{e^{\pi i/2}e^{-y}-e^{-\pi i/2}e^y}{2i}.$$ Now consider what $e^{\pi i/2}$ is, and you will be done.