:::section[From Two Basic Functions to Six] The functions $\sinh x$ and $\cosh x$ generate four quotient and reciprocal functions. Their identities follow from exponential algebra and from the relation between the even and odd parts of $e^x$. ::: :::definition[Quotient and Reciprocal Hyperbolic Functions] For real $x$ in the stated domains, $$\begin{align*} \tanh x&=\frac{\sinh x}{\cosh x} =\frac{e^x-e^{-x}}{e^x+e^{-x}},\\ \coth x&=\frac{\cosh x}{\sinh x},\qquad x\ne0,\\ \operatorname{sech}x&=\frac{1}{\cosh x},\\ \operatorname{csch}x&=\frac{1}{\sinh x},\qquad x\ne0. \end{align*}$$ Also, $\sinh0=0$ and $\cosh0=1$. ::: :::theorem[Addition and Subtraction Identities] For all real $x$ and $y$, $$\begin{align*} \sinh(x+y)&=\sinh x\cosh y+\cosh x\sinh y,\\ \sinh(x-y)&=\sinh x\cosh y-\cosh x\sinh y,\\ \cosh(x+y)&=\cosh x\cosh y+\sinh x\sinh y,\\ \cosh(x-y)&=\cosh x\cosh y-\sinh x\sinh y. \end{align*}$$ ::: :::proof[Proof] Using $e^u=\cosh u+\sinh u$ and $e^{-u}=\cosh u-\sinh u$, $$\begin{align*} e^{x+y}&=(\cosh x+\sinh x)(\cosh y+\sinh y),\\ e^{-(x+y)}&=(\cosh x-\sinh x)(\cosh y-\sinh y). \end{align*}$$ Subtracting these equations and dividing by $2$ yields the formula for $\sinh(x+y)$. Adding them and dividing by $2$ yields the formula for $\cosh(x+y)$. Replacing $y$ by $-y$ and using parity gives the subtraction formulas. $\square$ ::: :::corollary[Double-Angle and Exponential Identities] Setting $y=x$ gives $$\begin{align*} \sinh 2x&=2\sinh x\cosh x,\\ \cosh 2x&=\cosh^2x+\sinh^2x =2\cosh^2x-1 =2\sinh^2x+1. \end{align*}$$ Also, $$\begin{align*} \cosh x+\sinh x&=e^x,\\ \cosh x-\sinh x&=e^{-x}. \end{align*}$$ ::: :::theorem[Fundamental Hyperbolic Identity] For every real $x$, $$\begin{equation*} \cosh^2x-\sinh^2x=1. \end{equation*}$$ Consequently, $$\begin{align*} \operatorname{sech}^2x&=1-\tanh^2x,\\ \operatorname{csch}^2x&=\coth^2x-1,\qquad x\ne0. \end{align*}$$ ::: :::proof[Proof] Substituting the exponential definitions, $$\begin{align*} \cosh^2x-\sinh^2x &=\left(\frac{e^x+e^{-x}}{2}\right)^2 -\left(\frac{e^x-e^{-x}}{2}\right)^2\\ &=\frac{4e^xe^{-x}}{4}=1. \end{align*}$$ Dividing by $\cosh^2x$ and, where $x\ne0$, by $\sinh^2x$ gives the reciprocal identities. $\square$ ::: :::mistake[The Sign Differs from the Circular Identity] The corresponding trigonometric identity is $\cos^2x+\sin^2x=1$. For hyperbolic functions the correct sign is $$\begin{equation*} \cosh^2x-\sinh^2x=1. \end{equation*}$$ ::: :::exercise[Practice Questions] 1. Derive $\tanh(x+y)$ in terms of $\tanh x$ and $\tanh y$. 2. Express $\sinh^2x$ and $\cosh^2x$ in terms of $\cosh2x$. 3. Prove that $(\cosh x+\sinh x)(\cosh x-\sinh x)=1$. 4. Simplify $\operatorname{sech}^2x+\tanh^2x$. ::: :::answer[Answers and Guidance] 1. Divide the addition formulas for $\sinh(x+y)$ and $\cosh(x+y)$ by $\cosh x\cosh y$: $$\begin{equation*} \tanh(x+y)=\frac{\tanh x+\tanh y}{1+\tanh x\tanh y}. \end{equation*}$$ 2. $$\begin{equation*} \sinh^2x=\frac{\cosh2x-1}{2},\qquad \cosh^2x=\frac{\cosh2x+1}{2}. \end{equation*}$$ 3. The product equals $\cosh^2x-\sinh^2x=1$. 4. It equals $1$. ::: :::faq[Frequently Asked Questions] Q: Why is $\coth x$ undefined at zero? A: Its denominator is $\sinh0=0$, so the quotient is not defined. Q: Can the addition formulas be proved without memorizing them? A: Yes. Expand $e^{x+y}$ and $e^{-(x+y)}$, then take their half-sum and half-difference. Q: Why does $\operatorname{sech}^2x=1-\tanh^2x$ have a minus sign? A: It comes from dividing $\cosh^2x-\sinh^2x=1$ by $\cosh^2x$. ::: :::call-to-action[Continue Learning] subtitle: Interpret the fundamental identity geometrically and determine the domains and ranges of all six functions. button: Explore Geometry and Ranges|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/geometry-and-range-of-hyperbolic-functions button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::