From the definitions

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{\cosh x}& =\frac{1}{2}(e^x+e^{-x})\\ \multicolumn{1}{c}{\sinh x}& =\frac{1}{2}(e^x-e^{-x})\end{array}}$

we see that $\cosh x$ is always positive, it is an even function and its derivative is $\sinh x$ and $\sinh x$ is zero when $x=0$ so the graph of $\cosh x$ is U shaped with a minimum at $(0,1)$, as this is the only turning point and $\cosh x$ increases as $x$ increases for $x>0$.

The derivative of $\sinh x$ is $\cosh x$ so the graph of $\sinh x$ is everywhere increasing with the only turning point a point of inflexion at the origin. As $\sinh$ is an odd function the graph has rotational symmetry about the origin. Clearly for all $x$ we have $\sinh x<\cosh x$ so the graph of $\sinh$ is below the graph of $\cosh$ and they get closer as $x$ increases.

1. First, I have to prove that: ${\cosh }^{2}x-{\sinh }^{2}x=1$. From the definition $\cosh x+\sinh x=e^x$ and $\cosh x-\sinh x=e^{-x}$ so

${\displaystyle {\cosh }^{2}x-{\sinh }^{2}x=(\cosh x+\sinh x)(\cosh x-\sinh x)=e^x\times e^{-x}=1.}$

2. In the second part, I have to prove that: $\sinh 2x=2\sinh x\cosh x$.

I know that:

${\displaystyle \sinh 2x=\frac{1}{2}(e^{2x}-e^{-2x})}$

and

${\displaystyle 2\sinh x\cosh x=\frac{1}{2}(e^x-e^{-x})(e^x+e^{-x})=\frac{1}{2}(e^{2x}-e^{-2x}).}$

The two results above are equal, so I have proved that: $\sinh 2x=2\sinh x\cosh x$.

3. I have to prove that: $\sinh (n+1)x=\sinh \mathrm{nx}\cosh x+\cosh \mathrm{nx}\sinh x,$

Starting again from the definition, I can write the following expression for the terms:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{\sinh \mathrm{nx}\cosh x+\cosh \mathrm{nx}\sinh x}& =\frac{1}{4}[(e^{\mathrm{nx}}-e^{-\mathrm{nx}})(e^x+e^{-x})+(e^{\mathrm{nx}}+e^{-\mathrm{nx}})(e^x-e^{-x})]\\ \multicolumn{1}{c}{}& =\frac{1}{4}[2e^{(n+1)x}-2e^{-(n+1)x}]\\ \multicolumn{1}{c}{}& =\frac{1}{2}[e^{(n+1)x}-e^{-(n+1)x}]\\ \multicolumn{1}{c}{}& =\sinh (n+1)x.\end{array}}$

4. I have to prove the following inequality: $\sinh \mathrm{nx}\ge n\sinh x$.

This could be demonstrated using induction. It is clear that for $n=1$ I obtain an equality, so I look to prove the inequality for $n=2$. Then I shall consider it true for $n=k$ and prove it for $(k+1)$.

As $\cosh x>0$ for all $x$ it follows that $\sinh 2x=2\sinh x\cosh x>2\sinh x$ for all $x$ so the statement is true for $n=1$ and $n=2$. Now assume the statement is true for $n=k$ and consider $\sinh (k+1)x$. We have, as $\cosh$ is positive for all $x$,

${\displaystyle \sinh (k+1)x=\sinh \mathrm{kx}\cosh x+\cosh \mathrm{kx}\sinh x\ge \sinh \mathrm{kx}+\sinh x}$

and so if $\sinh \mathrm{kx}\ge k\sinh x$ then

${\displaystyle \sinh (k+1)x\ge k\sinh x+\sinh x=(k+1)\sinh x}$

and hence by the axiom of mathematical induction the statement is true for all positive integers $n$.