# Inverse Hyperbolic Functions

Hyperbolic cosine is y=cosh(x)=(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 (0,oo) then it will have unique inverse.

To find this inverse we use algorithm of finding inverse:

Let y=(e^x+e^(-x))/2.

Now, interchange x and y: x=(e^y+e^(-y))/2.

Finally, solve for y: e^y+e^(-y)=2x.

Multiply both sides of equation by e^y : e^(2y)+1=2xe^y.

Now, make substitution e^y=t then t^2-2xt+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(x+sqrt(x^2-1)).

Inverse Hyperbolic Cosine. y=text(arccosh)(x)=cosh^(-1)(x)=ln(x+sqrt(x^2-1)).

Hyperbolic sine is y=sinh(x)=(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=(e^x-e^(-x))/2.

Now, interchange x and y: x=(e^y-e^(-y))/2.

Finally, solve for y: e^y-e^(-y)=2x.

Multiply both sides of equation by e^y : e^(2y)-1=2xe^y.

Now, make substitution e^y=t then t^2-2xt-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(x+sqrt(x^2+1)).

Inverse Hyperbolic Sine. y=text(arcsinh)(x)=sinh^(-1)(x)=ln(x+sqrt(x^2+1)).

Hyperbolic tangent is y=tanh(x)=(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=(e^x-e^(-x))/(e^x+e^(-x)).

Now, interchange x and y: x=(e^y-e^(-y))/(e^y+e^(-y)).

Finally, solve for y: e^y-e^(-y)=x(e^y+e^(-y)).

Multiply both sides of equation by e^y : e^(2y)-1=x(e^(2y)+1).

From this we have that e^(2y)=(1+x)/(1-x) or y=1/2 ln((1+x)/(1-x)).

Inverse Hyperbolic Tangent. y=text(arctanh)(x)=tanh^(-1)(x)=1/2ln((1+x)/(1-x)).

Hyperbolic cotangent is y=coth(x)=(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=(e^x+e^(-x))/(e^x-e^(-x)).

Now, interchange x and y: x=(e^y+e^(-y))/(e^y-e^(-y)).

Finally, solve for y: e^y+e^(-y)=x(e^y-e^(-y)).

Multiply both sides of equation by e^y : e^(2y)+1=x(e^(2y)-1).

From this we have that e^(2y)=(x-1)/(x+1) or y=1/2 ln((x-1)/(x+1)).

Inverse Hyperbolic Cotangent. y= arccoth(x)=coth^(-1)(x)=1/2ln((x-1)/(x+1)).