Gosh Cosh
Explore the hyperbolic functions sinh and cosh using what you know about the exponential function.
Problem
The hyperbolic trig functions $\cosh $ and $\sinh $ are defined by $$\eqalign { \cosh x &= {1\over 2}(e^x + e^{-x}) \cr \sinh x &= {1\over 2}(e^x - e^{-x}).}$$ Using the definitions sketch the graphs of $\cosh x$ and $\sinh x$ on one diagram and prove the hyperbolic trig identities $$\eqalign { \cosh^2 x - \sinh^2 x &=1 \cr \sinh 2x &= 2\sinh x \cosh x \cr \sinh (n+1)x &= \sinh nx \cosh x + \cosh nx \sinh x.}$$
Notice the strong resemblance of these formulae to standard trigonometrical identities. Using this similarity as a guide, investigate the properties of a 'hyperbolic tangent' function $tanh(x)$ defined by
$$\tanh(x)=\frac{\sinh(x)}{\cosh(x)}$$
NOTES AND BACKGROUND
Notice that the identities for hyperbolic functions that you have proved are very similar to the ordinary trigonometric identities. In fact there is a complete hyperbolic geometry with similar results to the trigonometric results in Euclidean geometry. We compare absolute values in the corresponding result for $\sin nx$ which is $|\sin nx|\leq n|\sin x|$ . This formula needs the absolute values because the function is periodic and takes negative values for some multiples of the angle. Notice that the inequality in $|\sin nx|\leq n|\sin x|$ goes the other way to the corresponding hyperbolic result. This is because $\cos x \leq 1$ for all $x$ whereas $\cosh x\geq 1$.
Getting Started
To sketch the graphs, consider whether the functions are odd
or even and consider values of the functions for $x=0$ and for very
large $x$. Confirm your sketch by differentiating the functions and
finding the turning points.
To prove the hyperbolic trig identities requires only some
simple algebra.
Teachers' Resources
You do not need to have met hyperbolic trig functions before in order to do this question as the definitions are given and you are led through the question step by step. You could look up the results and the proofs of the hyperbolic trig identities but you will gain much more from thinking through these proofs for yourself.
Possible extension:
Prove, by induction or otherwise, that for $x> 0$ and positive integral values of $n> 1$ $$\sinh nx > n\sinh x.$$