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

Question

I need to show that (1) is true by letting $y = \sinh^{-1}x$ ...

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

... using (2) and (3) ...

$$\cosh^2(x) - \sinh^2(x) = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2\tag{2}$$

$$\cosh(x) + \sinh(x) = e^x\tag{3}$$

... with $x$ replaced by $y$.

However, I really am not sure how to jump from this ... : $$x = \frac{e^y - e^{-y}}{2}$$

... To anything related to these two equalities that I am supposed to use. Could someone please explain this one to me?

Edit: this question has already been answered before, but using quadratic equation. I wanted to prove the equality using the properties of hyperbolic functions alone

Answer 1

Here are the steps to prove that $\operatorname{sinh}^{-1}(x) = \ln(x + \sqrt{x^2 + 1})$ using the given equations:

1. Start with the expression $x = \frac{e^y - e^{-y}}{2}$, which relates $x$ and $y$ through the inverse hyperbolic sine function $\operatorname{sinh}^{-1}(x)$.

2. Substitute this expression for $x$ into the right-hand side of the equation to be proven: $ \ln(x + \sqrt{x^2 + 1}) $.

3. Simplify the expression inside the square root: $ x^2 + 1 = \left(\frac{e^y - e^{-y}}{2}\right)^2 + 1 = \frac{e^{2y} - 2 + e^{-2y}}{4} + 1 = \frac{e^{2y} + 2 + e^{-2y}}{4} $.

4. Use the equation $ \cosh^2(x) - \sinh^2(x) = 1 $ (derived from equation 2) to rewrite the expression inside the logarithm: $ \sqrt{x^2 + 1} = \sqrt{\frac{e^{2y} + 2 + e^{-2y}}{4}} = \frac{e^y + e^{-y}}{2} = \cosh(y) $.

5. Substitute this expression for $ \sqrt{x^2 + 1} $ into the right-hand side of the equation.

6. Use the equation $ \cosh(y) + \sinh(y) = e^y $ (derived from equation 3) to simplify the expression inside the logarithm: $ x + \sqrt{x^2 + 1} = \frac{e^y - e^{-y}}{2} + \cosh(y) = \sinh(y) + \cosh(y) = e^y $.

7. Simplify the right-hand side of the equation: $ \ln(e^y) = y $.

8. Therefore, $ \operatorname{sinh}^{-1}(x) = y = \ln(x + \sqrt{x^2 + 1}) $, which proves the original equation.

Answer 2

First note that statement 2 can be simplified:

$$\begin{align} \cosh^2(\alpha) - \sinh^2(\alpha) &= \left(\frac{e^\alpha + e^{-\alpha}}{2}\right)^2 - \left(\frac{e^\alpha - e^{-\alpha}}{2}\right)^2 \\[2mm] &= \left( \frac{e^{2\alpha} + 2 + e^{-2\alpha}}{4} \right) - \left( \frac{e^{2\alpha} -2 + e^{-2\alpha}}{4} \right) \\[2mm] &= 1 \end{align}$$

which leads to

2a.$\quad \cosh (\alpha) = \sqrt{\sinh^2 (\alpha) + 1}$

since $\cosh (\alpha) > 0$.

Now let $\sinh^{-1} x = y$; then $x = \sinh y$, and

$$\begin{align} \sinh^{-1} x &= y \\[2mm] &= \ln e^{y} \\[2mm] &= \ln \Bigl( \sinh (y) + \cosh (y) \Bigr) \\[2mm] &= \ln \left( \sinh (y) + \sqrt{\sinh^2 (y) + 1} \right) \\[2mm] &= \ln \left( x + \sqrt{x^2 + 1} \right) \end{align}$$

P.S. Proofs I've previously seen for this (including the ones in the link mentioned in the comments) have always resorted to solving a quadratic equation; I'm unclear why that seems to be the preferred method. If anyone can point out any advantages to that approach (or equivalently, any deficiencies in the approach I used) I would appreciate it.