Proof $ \cosh x=1 \Leftrightarrow x=0 $

Question

I just calculated the function value in $x=0$. Is that ok? If not, what am I suppose to do then?

Answer 1

It seems that you have proved that$$x=0\implies\cosh(x)=1.$$It remains to prove that$$\cosh(x)=1\implies x=0.$$In order to do that, use the fact that\begin{align}\cosh(x)=1&\iff e^x+e^{-x}=2\\&\iff e^{-x}(e^x-1)^2=0.\end{align}

Answer 2

Showing that $\cosh 0=1$ is one direction of the $\iff$ arrow. The other direction amounts to showing this:

No other real number $x$ has $\cosh x=1$.

There are several approaches to solve this, most conventionally through derivatives or the geometric definition of $\cosh$.

Answer 3

Hint

To prove $$\cosh x=1\iff x=0$$ is like to prove$$\exp x=1\iff x=0$$

Answer 4

No, that's not enough. I suggest the AM-GM inequality. since $\cosh x$ is the arithmetic mean of $e^{\pm x}$, whose geometric mean is $1$, it's $\ge1$ with equality iff $e^x=e^{-x}$, which is equivalent to $e^x=1$, or to $x=0$.