We know that $\cos hy=\frac{e^y+e^{-y}}{2} ...\left(i\right)$

Let $u=\frac{e^y+e^{-y}}{2},$ to find the maximum or minimum value of $u,$ find the points at which $\frac{du}{dy}=0.$

We have $u=\frac{e^y+e^{-y}}{2}$

$\Rightarrow \frac{du}{dy}=\frac{e^y-e^{-y}}{2}$

Now $\frac{du}{dy}=0$

$\Rightarrow \frac{e^y-e^{-y}}{2}=0$

$\Rightarrow e^y-\frac{1}{e^y}=0$

$\Rightarrow e^{2y}=1$

$\Rightarrow 2y=0$

$\Rightarrow y=0.$

Now, $\frac{d^{2}u}{d{y}^2}=\frac{e^y+e^{-y}}{2}>0$

We know that if the second derivative of a function is positive, then the function has a point of minima.

Hence, $u=\cos hy=\frac{e^y+e^{-y}}{2}$ is minimum at $y=0$ and the minimum value is

$\cos hy=\frac{e^0+e^0}{2}=\frac{1+1}{2}=1.$