4.1.2 Logarithmic Forms of Inverse Hyperbolic Functions

Understanding Inverse Hyperbolic Functions

The inverse hyperbolic functions ($\sinh^{-1} x, \cosh^{-1} x, \tanh^{-1} x$) can be expressed in logarithmic form. These expressions are useful for solving equations and evaluating inverse hyperbolic functions. The derivations rely on the definitions of the hyperbolic functions and the properties of exponential functions.

Deriving Logarithmic Forms

Inverse Hyperbolic Sine ($\sinh^{-1} x$):

Let $y = \sinh^{-1} x$, so $x = \sinh y$.

Using the definition of $\sinh y$:

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

Multiply through by $2$:

$$2x = e^y - e^{-y}$$

Rewrite as a quadratic in $e^y$:

$$(e^y)^2 - 2xe^y - 1 = 0$$

Solve using the quadratic formula:

$$e^y = x + \sqrt{x^2 + 1}$$

Take the natural logarithm:

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

Thus:

$$\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})$$

Inverse Hyperbolic Cosine ($\cosh^{-1} x$):

Let $y = \cosh^{-1} x$, so $x = \cosh y$

Using the definition of $\cosh y$:

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

Multiply through by $2$:

$$2x = e^y + e^{-y}$$

Rewrite as a quadratic in $e^y$:

$$(e^y)^2 - 2xe^y + 1 = 0$$

Solve using the quadratic formula:

$$e^y = x + \sqrt{x^2 - 1}$$

Take the natural logarithm:

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

Thus:

$$\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1}), \quad x \geq 1$$

Inverse Hyperbolic Tangent ($\tanh^{-1} x$):

Let $y = \tanh^{-1} x$, so $x = \tanh y$.

Using the definition of $\tanh y$:

$$x = \frac{\sinh y}{\cosh y} = \frac{e^y - e^{-y}}{e^y + e^{-y}}$$

Multiply through by $e^y + e^{-y}$:

$$x(e^y + e^{-y}) = e^y - e^{-y}$$

Combine terms:

$$(1 - x)e^y = (1 + x)e^{-y}$$

Divide through by $e^{-y}$:

$$(1 - x)e^{2y} = 1 + x$$

Solve for $e^{2y}$:

$$e^{2y} = \frac{1 + x}{1 - x}$$

Take the natural logarithm:

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

Thus:

$$\tanh^{-1} x = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right), \quad |x| < 1$$

Summary of Logarithmic Forms

$$\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})$$ $$\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1}), \quad x \geq 1$$ $$\tanh^{-1} x = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right), \quad |x| < 1$$

Worked Examples

Note Summary