H1 - Hyperbolic functions from exponentials

Defining sinh, cosh and tanh

Can the half-sum and half-difference of $3$ and $1/3$ both exceed 1, even though dividing the half-difference by the half-sum gives less than 1? The calculations $(3+1/3)/2$ and $(3-1/3)/2$ will give us a concrete starting point.

The numbers $3$ and $1/3$ are $e^x$ and $e^{-x}$ when $x=\ln3$. More generally, both exponentials are positive for every real $x$, but one grows while the other shrinks. Their half-difference and half-sum define two new functions.

Because the half-sum is positive, we may divide by it without excluding a real input. Hyperbolic tangent is

The last expression comes from multiplying numerator and denominator by $e^x$. For exact values at $x=\ln k$, use $e^x=k$ and $e^{-x}=1/k$, rather than rounding the logarithm first.

Here $e\approx2.71828$ is the base of natural logarithms, $\exp(x)$ is another notation for $e^x$, and $\ln x=\log_e x$ for $x>0$. More generally, $\log_a b=c$ means $a^c=b$, for $a>0$, $a\ne1$ and $b>0$.

Now change only the sign of the input. The two exponential values exchange places, so the difference changes sign while the sum stays the same.

These exact calculations also reveal a structural difference from circular trigonometry: hyperbolic sine and cosine can both exceed 1.

Try the same exponential calculation at a new logarithmic input.

Domains, ranges and graph shapes

All three functions accept every real input, but their possible outputs differ. A domain describes allowed inputs; a range describes the outputs actually reached.

Substituting $-x$ exchanges the exponentials. Therefore $\sinh(-x)=-\sinh x$ and $\tanh(-x)=-\tanh x$: these are odd functions, with rotational symmetry about the origin. In contrast, $\cosh(-x)=\cosh x$: it is even, with reflection symmetry in the vertical axis.

At zero, the definitions give $\sinh0=0$, $\cosh0=1$, $\tanh0=0$. To locate the minimum of cosh without calculus, observe that

Equality occurs only at $x=0$. Thus its range starts at 1, not at 0 or $-1$.

Why is the tangent range open at both ends? For positive $t=e^{2x}$, the expression $(t-1)/(t+1)$ is strictly between $-1$ and 1. As $x\to\infty$, divide by $e^{2x}$ to see the limit 1; as $x\to-\infty$, $e^{2x}\to0$, giving limit $-1$. No finite input attains either value. The exponentials also show that sinh grows without bound in both signed directions and that cosh grows without bound at both ends.