# Gosh Cosh

Explore the hyperbolic functions sinh and cosh using what you know about the exponential function.

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 $\cosh x\geq 1$.

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.

Image

Andrei Lazanu of Tudor Vianu National College, Bucharest, Romania sent in a very good solution to this problem.

From the definitions $$\eqalign { \cosh x &= {1\over 2}(e^x + e^{-x}) \cr \sinh x &= {1\over 2}(e^x - e^{-x})}$$ we see that $\cosh x$ is always positive, it is an even function and its derivative is $\sinh x$ and $\sinh x$ is zero when $x=0$ so the graph of $\cosh x$ is U shaped with a minimum at $(0,1)$, as this is the only turning point and $\cosh x$ increases as $x$ increases for $x> 0$.

The derivative of $\sinh x$ is $\cosh x$ so the graph of $\sinh x$ is everywhere increasing with the only turning point a point of inflexion at the origin. As $\sinh$ is an odd function the graph has rotational symmetry about the origin. Clearly for all $x$ we have $\sinh x < \cosh x$ so the graph of $\sinh$ is below the graph of $\cosh$ and they get closer as $x$ increases.

1. First, I have to prove that: $\cosh^2 x - \sinh^2 x =1$. From the definition $\cosh x + \sinh x = e^x$ and $\cosh x - \sinh x = e^{-x}$ so $$\cosh^2 x - \sinh^2 x = (\cosh x + \sinh x)(\cosh x - \sinh x) = e^x \times e^{-x} = 1.$$ 2. In the second part, I have to prove that: $\sinh 2x = 2\sinh x \cosh x$.

I know that: $$\sinh 2x = {1\over 2}(e^{2x} - e^{-2x})$$ and $$2\sinh x \cosh x = {1\over 2}(e^x - e^{-x})(e^x + e^{-x})={1\over 2 }(e^{2x}-e^{-2x}).$$ The two results above are equal, so I have proved that: $\sinh 2x = 2\sinh x \cosh x$.

3. I have to prove that:$\sinh(n+1)x = \sinh nx \cosh x + \cosh nx \sinh x$

Starting again from the definition, I can write the following expression for the terms: $$\eqalign{ \sinh nx \cosh x + \cosh nx \sinh x &= {1\over 4}[(e^{nx} - e^{-nx})(e^x + e^{-x})+ (e^{nx} + e^{-nx})(e^x - e^{-x})] \cr &= {1\over 4}[2e^{(n+1)x}-2e^{-(n+1)x}] \cr &= {1\over 2}[e^{(n+1)x} - e^{-(n+1)x}] \cr &= \sinh (n+1)x}$$

4. I have to prove the following inequality: $\sinh nx \geq n \sinh x$.

This could be demonstrated using induction. It is clear that for $n = 1$ I obtain an equality, so I look to prove the inequality for $n = 2$. Then I shall consider it true for $n=k$ and prove it for $(k+1)$.

As $\cosh x > 0$ for all $x$ it follows that $\sinh 2x = 2\sinh x \cosh x > 2\sinh x$ for all $x$ so the statement is true for $n=1$ and $n=2$. Now assume the statement is true for $n=k$ and consider $\sinh(k+1)x$. We have, as $\cosh$ is positive for all $x$, $$\sinh(k+1)x = \sinh kx \cosh x + \cosh kx \sinh x \geq \sinh kx + \sinh x$$ and so if $\sinh kx \geq k\sinh x$ then $$\sinh(k+1)x \geq k\sinh x + \sinh x = (k+1)\sinh x$$ and hence by the axiom of mathematical induction the statement is true for all positive integers $n$.

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 $n> 1$ $$\sinh nx > n\sinh x.$$