Inverse Hyperbolic Functions

Hyperbolic cosine is $$${y}={\cosh{{\left({x}\right)}}}=\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}{{2}}$$$.

This function is not one-to-one, so there is no unique inverse for this function. However, if we take function on interval $$${\left({0},\infty\right)}$$$ then it will have unique inverse.

To find this inverse we use algorithm of finding inverse:

Let $$${y}=\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}{{2}}$$$.

Now, interchange $$${x}$$$ and $$${y}$$$: $$${x}=\frac{{{{e}}^{{y}}+{{e}}^{{-{y}}}}}{{2}}$$$.

Finally, solve for y: $$${{e}}^{{y}}+{{e}}^{{-{y}}}={2}{x}$$$.

Multiply both sides of equation by $$${{e}}^{{y}}$$$ : $$${{e}}^{{{2}{y}}}+{1}={2}{x}{{e}}^{{y}}$$$.

Now, make substitution $$${{e}}^{{y}}={t}$$$ then $$${{t}}^{{2}}-{2}{x}{t}+{1}={0}$$$. This is quadratic equation.

It has two solutions: $$${t}_{{1}}={x}+\sqrt{{{{x}}^{{2}}-{1}}}$$$ and $$${t}_{{2}}={x}-\sqrt{{{{x}}^{{2}}-{1}}}$$$.

So, either $$${{e}}^{{y}}={x}+\sqrt{{{{x}}^{{2}}-{1}}}$$$ or $$${{e}}^{{y}}={x}-\sqrt{{{{x}}^{{2}}-{1}}}$$$.

Since we took $$${x}>{0}$$$ and interchanged $$${x}$$$ and $$${y}$$$ then we require $$${y}>{0}$$$, so only $$${{e}}^{{y}}={x}+\sqrt{{{{x}}^{{2}}-{1}}}$$$ is applicable. From this we have that $$${y}={\ln{{\left({x}+\sqrt{{{{x}}^{{2}}-{1}}}\right)}}}$$$.

Hyperbolic sine is $$${y}={\sinh{{\left({x}\right)}}}=\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{2}}$$$.

This function is one-to-one, therefore it has unique inverse.

To find this inverse we use algorithm of finding inverse:

Let $$${y}=\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{2}}$$$.

Now, interchange $$${x}$$$ and $$${y}$$$: $$${x}=\frac{{{{e}}^{{y}}-{{e}}^{{-{y}}}}}{{2}}$$$.

Finally, solve for y: $$${{e}}^{{y}}-{{e}}^{{-{y}}}={2}{x}$$$.

Multiply both sides of equation by $$${{e}}^{{y}}$$$ : $$${{e}}^{{{2}{y}}}-{1}={2}{x}{{e}}^{{y}}$$$.

Now, make substitution $$${{e}}^{{y}}={t}$$$ then $$${{t}}^{{2}}-{2}{x}{t}-{1}={0}$$$. This is quadratic equation.

It has two solutions: $$${t}_{{1}}={x}+\sqrt{{{{x}}^{{2}}+{1}}}$$$ and $$${t}_{{2}}={x}-\sqrt{{{{x}}^{{2}}+{1}}}$$$.

But $$${t}_{{2}}<{0}$$$ for all $$${x}$$$ and $$${t}={{e}}^{{y}}$$$ should be positive, therefore, $$${t}_{{1}}={x}+\sqrt{{{{x}}^{{2}}+{1}}}$$$.

This gives us that $$${{e}}^{{y}}={x}+\sqrt{{{{x}}^{{2}}+{1}}}$$$. From this we have that $$${y}={\ln{{\left({x}+\sqrt{{{{x}}^{{2}}+{1}}}\right)}}}$$$.

Hyperbolic tangent is $$${y}={\tanh{{\left({x}\right)}}}=\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}$$$.

This function is one-to-one, therefore it has unique inverse.

To find this inverse we use algorithm of finding inverse:

Let $$${y}=\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}$$$.

Now, interchange $$${x}$$$ and $$${y}$$$: $$${x}=\frac{{{{e}}^{{y}}-{{e}}^{{-{y}}}}}{{{{e}}^{{y}}+{{e}}^{{-{y}}}}}$$$.

Finally, solve for y: $$${{e}}^{{y}}-{{e}}^{{-{y}}}={x}{\left({{e}}^{{y}}+{{e}}^{{-{y}}}\right)}$$$.

Multiply both sides of equation by $$${{e}}^{{y}}$$$ : $$${{e}}^{{{2}{y}}}-{1}={x}{\left({{e}}^{{{2}{y}}}+{1}\right)}$$$.

From this we have that $$${{e}}^{{{2}{y}}}=\frac{{{1}+{x}}}{{{1}-{x}}}$$$ or $$${y}=\frac{{1}}{{2}}{\ln{{\left(\frac{{{1}+{x}}}{{{1}-{x}}}\right)}}}$$$.

Hyperbolic cotangent is $$${y}={\coth{{\left({x}\right)}}}=\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}$$$.

This function is one-to-one, therefore it has unique inverse.

To find this inverse we use algorithm of finding inverse:

Let $$${y}=\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}$$$.

Now, interchange $$${x}$$$ and $$${y}$$$: $$${x}=\frac{{{{e}}^{{y}}+{{e}}^{{-{y}}}}}{{{{e}}^{{y}}-{{e}}^{{-{y}}}}}$$$.

Finally, solve for $$${y}$$$: $$${{e}}^{{y}}+{{e}}^{{-{y}}}={x}{\left({{e}}^{{y}}-{{e}}^{{-{y}}}\right)}$$$.

Multiply both sides of equation by $$${{e}}^{{y}}$$$ : $$${{e}}^{{{2}{y}}}+{1}={x}{\left({{e}}^{{{2}{y}}}-{1}\right)}$$$.

From this we have that $$${{e}}^{{{2}{y}}}=\frac{{{x}-{1}}}{{{x}+{1}}}$$$ or $$${y}=\frac{{1}}{{2}}{\ln{{\left(\frac{{{x}-{1}}}{{{x}+{1}}}\right)}}}$$$.