:::section[Why Branch Restrictions Matter] An inverse function exists only after the original function is one-to-one on the chosen domain. The functions $\sinh$ and $\tanh$ are already one-to-one on $\mathbb{R}$. The even function $\cosh$ is restricted to $[0,\infty)$. ::: :::definition[Principal Inverse Hyperbolic Functions] The principal inverse functions are defined by the following branch choices: $$\begin{align*} \sinh^{-1}&:\mathbb{R}\to\mathbb{R},\\ \cosh^{-1}&:[1,\infty)\to[0,\infty),\\ \tanh^{-1}&:(-1,1)\to\mathbb{R}. \end{align*}$$ Here the superscript $-1$ denotes an inverse function, not a reciprocal. ::: :::theorem[Logarithmic Forms] On their real domains, $$\begin{align*} \sinh^{-1}x&=\ln\left(x+\sqrt{x^2+1}\right), &&x\in\mathbb{R},\\ \cosh^{-1}x&=\ln\left(x+\sqrt{x^2-1}\right), &&x\ge1,\\ \tanh^{-1}x&=\frac12\ln\left(\frac{1+x}{1-x}\right), &&|x|<1. \end{align*}$$ ::: :::proof[Derivation of the Inverse Sine Form] Let $y=\sinh^{-1}x$. Then $x=\sinh y$, so $$\begin{equation*} 2x=e^y-e^{-y}. \end{equation*}$$ Multiplying by $e^y$ gives $$\begin{equation*} (e^y)^2-2xe^y-1=0. \end{equation*}$$ Solving this quadratic, $$\begin{equation*} e^y=x\pm\sqrt{x^2+1}. \end{equation*}$$ Since $e^y>0$ and $x+\sqrt{x^2+1}>0$, the positive admissible root is $$\begin{equation*} e^y=x+\sqrt{x^2+1}. \end{equation*}$$ Taking logarithms yields the formula. $\square$ ::: :::proof[Derivation of the Inverse Cosine and Tangent Forms] For $y=\cosh^{-1}x$ with $y\ge0$, solving $$\begin{equation*} x=\frac{e^y+e^{-y}}2 \end{equation*}$$ gives $e^y=x+\sqrt{x^2-1}$, because $e^y\ge1$. Hence the stated logarithm follows. For $y=\tanh^{-1}x$, $$\begin{align*} x&=\frac{e^{2y}-1}{e^{2y}+1},\\ e^{2y}(1-x)&=1+x. \end{align*}$$ Since $|x|<1$, both $1+x$ and $1-x$ are positive. Therefore $$\begin{equation*} y=\frac12\ln\left(\frac{1+x}{1-x}\right). \end{equation*}$$ $\square$ ::: :::key-formula[Other Principal Logarithmic Forms] $$\begin{align*} \coth^{-1}x &=\frac12\ln\left(\frac{x+1}{x-1}\right), &&|x|>1,\\ \operatorname{sech}^{-1}x &=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right), &&0<x\le1,\\ \operatorname{csch}^{-1}x &=\ln\left(\frac1x+\frac{\sqrt{1+x^2}}{|x|}\right), &&x\ne0. \end{align*}$$ ::: :::mistake[Check the Logarithm Argument and the Branch] A formally obtained quadratic root is not automatically admissible. The selected root must be positive because it equals $e^y$, and the chosen inverse branch determines the sign of $y$. ::: :::exercise[Practice Questions] 1. Evaluate $\sinh^{-1}(0)$, $\cosh^{-1}(1)$, and $\tanh^{-1}(0)$. 2. Show that $\sinh^{-1}x$ is an odd function. 3. Express $\sinh^{-1}(3/4)$ as a logarithm and simplify. 4. State the real domain of $\operatorname{sech}^{-1}x$. ::: :::answer[Answers and Guidance] 1. All three values are $0$. 2. Use the logarithmic form and the identity $$\begin{equation*} \left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)=1. \end{equation*}$$ It follows that $\sinh^{-1}(-x)=-\sinh^{-1}x$. 3. $$\begin{equation*} \sinh^{-1}\left(\frac34\right) =\ln\left(\frac34+\frac54\right)=\ln2. \end{equation*}$$ 4. The domain is $(0,1]$. ::: :::faq[Frequently Asked Questions] Q: Why is $\cosh$ restricted to $[0,\infty)$? A: It is even and therefore not one-to-one on all of $\mathbb{R}$. On $[0,\infty)$ it is strictly increasing and has range $[1,\infty)$. Q: Why is the domain of $\tanh^{-1}x$ equal to $(-1,1)$? A: That interval is the range of $\tanh x$ on $\mathbb{R}$. Q: Are the logarithmic formulas valid outside the stated real domains? A: Not as real-valued principal inverse functions. Extensions to complex values require additional branch conventions. ::: :::call-to-action[Continue Learning] subtitle: Apply the branch choices to the two possible orders of composing a hyperbolic function with its inverse. button: Study Composition Laws|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/composition-laws-for-inverse-hyperbolic-functions button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::