Hyperbolic functions and their graphsAQA A-Level Further Maths: Revision notes

The hyperbolic functions are built from the exponential function: $\sinh x=\frac{e^x-e^{-x}}{2},\qquad\cosh x=\frac{e^x+e^{-x}}{2},\qquad\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}.$ They are read as 'shine', 'cosh' and 'than'. At $x=0$: $\sinh0=0$, $\cosh0=1$, $\tanh0=0$. Example, $x=\ln3$ (so $e^x=3$ and $e^{-x}=\frac13$): $\sinh x=\frac12\left(3-\frac13\right)=\frac43$, $\cosh x=\frac53$ and $\tanh x=\frac45$.

Since $\sinh(-x)=-\sinh x$, the function is odd, so the graph has rotational symmetry of order 2 about the origin. It passes through $(0,0)$ and increases everywhere. For large positive $x$, $e^{-x}$ is tiny so $\sinh x\approx\frac12e^x$; for large negative $x$, $\sinh x\approx-\frac12e^{-x}$. The curve therefore rises steeply on the right and falls steeply on the left, like a stretched cubic, with no turning points and no asymptotes.

Since $\cosh(-x)=\cosh x$, the function is even and the graph is symmetrical about the $y$-axis. It has a minimum at $(0,1)$, because $\frac12(e^x+e^{-x})\ge1$ with equality only when $x=0$. For large $|x|$, $\cosh x\approx\frac12e^{|x|}$, so the curve climbs steeply either side of the minimum. It never crosses the $x$-axis. The shape is called a catenary, the curve of a hanging chain, and it is not a parabola.

$\tanh x$ is odd, passes through the origin and increases everywhere. Writing $\tanh x=\frac{1-e^{-2x}}{1+e^{-2x}}$, as $x\to\infty$ we have $e^{-2x}\to0$ so $\tanh x\to1$; similarly $\tanh x\to-1$ as $x\to-\infty$. The graph therefore has two horizontal asymptotes, $y=1$ and $y=-1$, is steepest at the origin, and is S-shaped. It stays strictly between $-1$ and $1$ and is defined for all real $x$: there are no vertical asymptotes.

On one set of axes, $\cosh x$ always lies above $\sinh x$, because $\cosh x-\sinh x=e^{-x}>0$. The gap shrinks to 0 as $x\to\infty$, so for large positive $x$ both curves lie close to $y=\frac12e^x$. Key features to mark on a sketch: the intercept $(0,1)$ for $\cosh$; the origin for $\sinh$ and $\tanh$; the asymptotes $y=\pm1$ for $\tanh$; the symmetry of each curve. To solve equations such as $2\cosh x+3\sinh x=3$, replace each function by its exponential form and let $u=e^x$ to obtain a quadratic in $u$. Reject any root with $u\le0$, since $e^x>0$.