CP 8.1 - The functions sinh, cosh and tanh

Three functions made from exponentials

Can an equation such as $\sinh 2x=4\sinh x$ have three real solutions even though the exponential function is one-to-one? It can: the new functions combine exponentials in different ways. You will find all three solutions at the end.

Take $x=\ln3$. Then $\mathrm{e}^x=3$ and $\mathrm{e}^{-x}=1/3$. Their half-sum is $5/3$; their half-difference is $4/3$. These two combinations are the starting point.

The names hyperbolic sine, hyperbolic cosine and hyperbolic tangent are abbreviated as follows. Read $\sinh x$ as one function, not as a product of letters.

The denominator of $\tanh x$ is never zero: $\cosh x$ is the half-sum of two positive numbers. Thus all three functions are defined for every real input.

Change only the sign of the input. The two exponentials exchange places, so their difference reverses sign but their sum does not.

These symmetries will give two of the graphs rotational symmetry about the origin and the other reflection symmetry in the vertical axis.

Use a different logarithmic input without rounding.

Graphs, domains and ranges

The exponential definitions determine the shapes of the graphs, including their symmetry, turning points and asymptotes.

For $\sinh x$, increasing $x$ increases $\mathrm{e}^x$ and decreases $\mathrm{e}^{-x}$. Subtracting the latter makes both changes increase the result. Thus $\sinh x$ is strictly increasing, passes through the origin, and takes every real value: it grows without bound to the right and becomes arbitrarily negative to the left.

For $\cosh x$, the symmetry is even. Its smallest value is $1$, because

Equality holds only at $x=0$. The function decreases towards $(0,1)$ on the left and increases away from it on the right. It never crosses either axis except for its vertical intercept $(0,1)$.

For $\tanh x$, divide the defining numerator and denominator by $\mathrm{e}^{-x}$:

As $x$ increases, the last fraction decreases, so $\tanh x$ increases. It approaches $1$ as $x$ tends to infinity and $-1$ as $x$ tends to negative infinity, but reaches neither at a finite input.

[DIAGRAM: asset_name: The three hyperbolic graphs; asset_slug: 9fm0-hyperbolic-graphs; caption: All three domains are real, but their ranges and symmetries differ; description: Three separate labelled Cartesian panels show sinh x increasing through the origin with odd symmetry, cosh x with even symmetry and minimum (0,1), and tanh x increasing through the origin between dashed horizontal asymptotes y equals minus 1 and y equals 1. Equal positive and negative horizontal extents are used, and each panel states its range.]

None of these graphs is periodic. Despite the familiar-looking names, copying the oscillating shapes of sine and cosine would give the wrong functions entirely. Their definitions and graph features are not supplied in the formula booklet; learn them.

Use those ranges and symmetries to count solutions before calculating any.

Identities from the definitions

Hyperbolic identities can be proved by substituting the exponential definitions and collecting terms; their signs are part of the mathematics, not a detail to borrow from trigonometry.

For the fundamental identity,

The cross-terms survive with a plus sign after the subtraction. It follows that $1-\tanh^2x=1/\cosh^2x$, since $\cosh x$ never vanishes. The reciprocal $1/\cosh x$ is written $\operatorname{sech}x$, so this is also $1-\tanh^2x=\operatorname{sech}^2x$.