A calculus of the absurd

15.4 Identities

In general, all the trigonometric identities also hold for hyperbolic functions, albeit with minor modifications.

The basic rule of thumb is that wherever you see \(\pm \sin ^2(x)\) in "normal" trigonometry, replace it with \(\mp \sinh ^2(x)\) (so \(\sin ^2(x)\) would be replaced with \(-\sinh ^2(x)\), and \(-\sin ^2(x)\) with \(\sinh ^2(x)\)). This comes from the relationship \(\sin (ix) = i\sinh (x)\) which means that \(\sin ^2(ix) = i^2 \sinh ^2(x) = -\sinh ^2(x)\). Why this is true was explored in the previous section.

To prove identities, the usual thing to try is to write everything in terms of the exponential function and go from there.

Example 15.4.1 Show that
\(\seteqnumber{0}{15.}{33}\)
\begin{equation*} \cosh (2x) = \cosh ^2(x) + \sinh ^2(x) \end{equation*}

Solution: Usually when proving identities it’s easiest to start with the more "complicated" ¹¹⁷¹¹⁷ This is a purely qualitative distinction, but it’s usually the side that makes you think either "yuck" or "what a fun challenge" depending on your view of mathematics. side.

\begin{align} \cosh ^2(x) + \sinh ^2(x) & = \left ( \frac {e^{2x}+1}{2e^x} \right ) ^2 + \left ( \frac {e^{2x}-1}{2e^x} \right ) ^2 \label {cosh2x bracket expansion} \\ & = \frac {2e^{4x} + 2}{4e^{2x}} \\ & = \frac {e^{4x}+1}{2e^{2x}} \\ & = \cosh (2x) \end{align}

Note that for the binomials in Equation 15.34 we know that the middle terms will cancel because the brackets are the same, except for a \(-1\) in one and a \(1\) in the other.