:::section[Why Hyperbolic Functions Arise] The exponential function contains an even part and an odd part. Given that $$\begin{equation*} e^x=\frac{e^x+e^{-x}}{2}+\frac{e^x-e^{-x}}{2}, \end{equation*}$$ the first summand is unchanged when $x$ is replaced by $-x$, whereas the second changes sign. These two parts define the basic hyperbolic functions. ::: :::definition[Hyperbolic Sine and Hyperbolic Cosine] For every $x\in\mathbb{R}$, $$\begin{align*} \sinh x&=\frac{e^x-e^{-x}}{2},\\ \cosh x&=\frac{e^x+e^{-x}}{2}. \end{align*}$$ The domain of each function is $\mathbb{R}$. Since $\sinh x$ is continuous, strictly increasing, and unbounded in both directions, its range is $\mathbb{R}$. Since $e^x+e^{-x}\ge 2$, the range of $\cosh x$ is $[1,\infty)$. ::: :::theorem[Parity of Hyperbolic Sine and Cosine] For every $x\in\mathbb{R}$, $$\begin{align*} \sinh(-x)&=-\sinh x,\\ \cosh(-x)&=\cosh x. \end{align*}$$ Thus $\sinh$ is odd and $\cosh$ is even. ::: :::proof[Proof] Substituting $-x$ into the definitions gives $$\begin{align*} \sinh(-x) &=\frac{e^{-x}-e^x}{2} =-\frac{e^x-e^{-x}}{2} =-\sinh x,\\ \cosh(-x) &=\frac{e^{-x}+e^x}{2} =\frac{e^x+e^{-x}}{2} =\cosh x. \end{align*}$$ Therefore the stated parity properties follow. $\square$ ::: :::example[Basic Values and Graph Information] Substituting $x=0$ gives $\sinh 0=0$ and $\cosh 0=1$. Hence the graph of $\sinh x$ passes through the origin and has origin symmetry. The graph of $\cosh x$ passes through $(0,1)$ and has symmetry about the $y$-axis. Moreover, $$\begin{equation*} \cosh x-\sinh x=e^{-x}>0, \end{equation*}$$ so $\cosh x>\sinh x$ for every real $x$. ::: :::mistake[Do Not Confuse the Two Symmetries] Oddness means $f(-x)=-f(x)$ and gives origin symmetry. Evenness means $f(-x)=f(x)$ and gives $y$-axis symmetry. The two statements are not interchangeable. ::: :::exercise[Practice Questions] 1. Use the definitions to compute $\sinh(\ln 2)$ and $\cosh(\ln 2)$ exactly. 2. Prove directly that $\cosh x\ge 1$ for every real $x$, with equality only at $x=0$. 3. Determine whether $\sinh x+\cosh x$ is even, odd, or neither. 4. Show that $e^x=\cosh x+\sinh x$. ::: :::answer[Answers and Guidance] 1. Since $e^{\ln 2}=2$ and $e^{-\ln 2}=1/2$, $$\begin{equation*} \sinh(\ln 2)=\frac{3}{4},\qquad \cosh(\ln 2)=\frac{5}{4}. \end{equation*}$$ 2. By the arithmetic-geometric mean inequality, $e^x+e^{-x}\ge 2$, so $\cosh x\ge1$. Equality requires $e^x=e^{-x}$, hence $x=0$. 3. Since $\sinh$ is odd and $\cosh$ is even, their sum is generally neither even nor odd. 4. Adding the definitions cancels the $e^{-x}$ terms and gives $e^x$. ::: :::faq[Frequently Asked Questions] Q: Why are these functions called hyperbolic? A: The point $(\cosh t,\sinh t)$ lies on the unit hyperbola $x^2-y^2=1$, just as $(\cos t,\sin t)$ lies on the unit circle. Q: Is $\cosh x$ ever zero? A: No. Since $e^x>0$ and $e^{-x}>0$, their average is positive; in fact $\cosh x\ge1$. Q: Does the superscript in $\sinh^{-1}x$ mean a reciprocal? A: In inverse-function notation it means the inverse function, not $1/\sinh x$. The reciprocal is denoted $\operatorname{csch}x$. ::: :::call-to-action[Continue Learning] subtitle: Use the exponential construction to derive the algebraic identities that organize all six hyperbolic functions. button: Study Fundamental Identities|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/fundamental-identities-of-hyperbolic-functions button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::