# 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 ${y}={\ln{{\left({x}+\sqrt{{{{x}}^{{2}}-{1}}}\right)}}}$.

Inverse Hyperbolic Cosine. ${y}=\text{arccosh}{\left({x}\right)}={{\cosh}}^{{-{1}}}{\left({x}\right)}={\ln{{\left({x}+\sqrt{{{{x}}^{{2}}-{1}}}\right)}}}$.

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 ${y}={\ln{{\left({x}+\sqrt{{{{x}}^{{2}}+{1}}}\right)}}}$.

Inverse Hyperbolic Sine. ${y}=\text{arcsinh}{\left({x}\right)}={{\sinh}}^{{-{1}}}{\left({x}\right)}={\ln{{\left({x}+\sqrt{{{{x}}^{{2}}+{1}}}\right)}}}$.

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 ${y}=\frac{{1}}{{2}}{\ln{{\left(\frac{{{1}+{x}}}{{{1}-{x}}}\right)}}}$.

Inverse Hyperbolic Tangent. ${y}=\text{arctanh}{\left({x}\right)}={{\tanh}}^{{-{1}}}{\left({x}\right)}=\frac{{1}}{{2}}{\ln{{\left(\frac{{{1}+{x}}}{{{1}-{x}}}\right)}}}$.

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 ${y}=\frac{{1}}{{2}}{\ln{{\left(\frac{{{x}-{1}}}{{{x}+{1}}}\right)}}}$.

Inverse Hyperbolic Cotangent. ${y}={\operatorname{arccoth}{{\left({x}\right)}}}={{\coth}}^{{-{1}}}{\left({x}\right)}=\frac{{1}}{{2}}{\ln{{\left(\frac{{{x}-{1}}}{{{x}+{1}}}\right)}}}$.