$\large \cosh^{-1}{x} \,=\, \log_{e}{(x+\sqrt{x^2-1})}$

The inverse form of the hyperbolic cosine function is called the inverse hyperbolic cosine function.

The hyperbolic cosine function is defined in exponential functions form. Hence, the inverse hyperbolic cosine function should be in logarithmic function form and it can be derived mathematically from hyperbolic cosine function.

Take two literals $x$ and $y$. Assume, the value of $x$ is equal to the hyperbolic cosine of $y$ and it is written in mathematical form as follows.

$x \,=\, \cosh{y}$

According to this relation, the value of $y$ should be equal to the inverse cosine of $x$ and it is written in the following way.

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

The relation between them can be expressed in mathematical form as follows.

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

01.

On the basis of definition of hyperbolic cosine function, hyperbolic cosine function can be expressed in terms of exponential functions as follows.

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

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

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

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

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

$\implies 2xe^y = e^{2y}+1$

$\implies 0 = e^{2y}-2xe^y+1$

$\implies e^{2y}-2xe^y+1=0$

$\implies {(e^y)}^2-2xe^y+1=0$

Take $u = e^y$ and change the equation in terms of $u$.

$\,\,\, \therefore \,\,\,\,\,\, u^2-2xu+1=0$

02.

The quadratic equation is in terms of $u$ and solve it to get the value of $u$.

$u$ $\,=\,$ $\dfrac{-(-2x) \pm \sqrt{{(-2x)}^2-4 \times 1 \times 1}}{2 \times 1}$

$\implies u = \dfrac{2x \pm \sqrt{4x^2-4}}{2}$

$\implies u = \dfrac{2x \pm \sqrt{4(x^2-1)}}{2}$

$\implies u = \dfrac{2x \pm 2\sqrt{x^2-1}}{2}$

$\implies u = \dfrac{2(x \pm \sqrt{x^2-1})}{2}$

$\implies \require{cancel} u = \dfrac{\cancel{2}(x \pm \sqrt{x^2-1})}{\cancel{2}}$

$\implies u = x \pm \sqrt{x^2-1}$

In fact $u = e^y$. So, replace $u$ by $e^y$ in the solution.

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

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

03.

The value of the mathematical constant $e$ is a positive irrational number but the value of $y$ can be either a positive or a negative real number. Hence, the value of $e$ raised to the power of y should be a positive number. In other words, $e^y > 0$.

So, verify both solutions of $e^y$ by substituting real numbers instead of $x$.

Therefore, $x+\sqrt{x^2-1} > 0$ and $x-\sqrt{x^2-1} > 0$

Observe both inequations, the value of $x$ should be great than zero. Test both inequations for positive values.

If $x = 1$, then $1+\sqrt{{(1)}^2-1} = 1$ and $1-\sqrt{{(1)}^2-1} = 1$

If $x = 2$, then $2+\sqrt{{(2)}^2-1} = 3.7321$ and $2-\sqrt{{(2)}^2-1} = 0.2679$

If $x = 3$, then $3+\sqrt{{(3)}^2-1} = 5.8284$ and $3-\sqrt{{(3)}^2-1} = 0.1716$

If $x = 4$, then $4+\sqrt{{(4)}^2-1} = 7.8730$ and $4-\sqrt{{(4)}^2-1} = 0.1270$

$\vdots$

As the value of $x$ is increased, the value of first inequation is increased and giving a positive value whereas the value of second inequation is also giving positive value but there is a decrement in its value. So, the second inequation surely gives negative values at some point and it violates the rule.

Hence, $e^y = x+\sqrt{x^2-1}$ and $e^y \ne x-\sqrt{x^2-1}$

04.

Take natural logarithm both sides.

$\implies \log_{e}{e^y} = \log_{e}{(x+\sqrt{x^2-1})}$

$\implies y \times \log_{e}{e} = \log_{e}{(x+\sqrt{x^2-1})}$

$\implies y \times 1 = \log_{e}{(x+\sqrt{x^2-1})}$

$\implies y = \log_{e}{(x+\sqrt{x^2-1})}$

As per our early assumption, $y = \cosh^{-1}{x}$

$\therefore \,\,\,\,\,\, \cosh^{-1}{x} = \log_{e}{(x+\sqrt{x^2-1})}$

It can be simply written as follows.

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

List of most recently solved mathematics problems.

Jun 22, 2018

Integral Calculus

Evaluate $\displaystyle \int \dfrac{1+\cos{4x}}{\cot{x}-\tan{x}} dx$

Jun 21, 2018

Limit

Evaluate $\displaystyle \large \lim_{x \to \infty} \normalsize {\sqrt{x^2+x+1}-\sqrt{x^2+1}}$

Jun 20, 2018

Differentiation

Learn how to find derivative of $\sin{(x^2)}$ with respect to $x$.

Jun 19, 2018

Limit (Calculus)

Find $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{x-\sin{x}}{x^3}$

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.