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))`.