1.5 Trigonometric and Hyperbolic Functions

1.5.1 Definitions

We recall Euler’s formula for the sine and cosine,

exp(ix)=cos(x)+isin(x)

where x is a real number. From this, we can express sine and cosine as

cos(x)sin(x):=:=21eix+e−ix12ieix−e−ix

We now define the hyperbolic functions ‘hyperbolic cosine’ and ‘hyperbolic sine’ as

cosh(x)sinh(x):=:=21ex+e−x21ex−e−x

i.e. analogous to cosine and sine but without the imaginary unit i. Using i2=−1, we recognise that

cosh(x)=cos(ix)sinh(x)=−isin(ix)

which means that trigonometric and hyperbolic functions are closely related. Their behaviour as a function of x, however, is different: while sine and cosine are oscillatory functions, the hyperbolic functions cosh(x) and sinh(x) are not oscillatory, because they are just linear combinations of ex and e−x which are not oscillatory. We have the following properties:

cosh(0)cosh(x→∞)sinh(0)sinh(x→∞)=→=→1cosh(x)=cosh(−x)21ex→x→∞∞cosh(x→−∞)→21e−x→x→−∞∞0sinh(x)=−sinh(−x)21ex→x→∞∞sinh(x→−∞)→−21e−x→x→−∞−∞

from which we already can sketch the two hyperbolic functions, see Fig. 1.2.

In addition, one defines the hyperbolic tangent and cotangent

tanh(x):=sinh(x)cosh(x)coth(x):=sinh(x)cosh(x)

1.5.2 Inverse hyperbolic functions

Inverting

y=sinh(x)→x=sinh−1(y)

we find the inverse hyperbolic sine sinh−1 by setting

y=2ex−e−x⇝e2x−2yex−1=0

This is a quadratic equation in u=ex with the solutions

u±=y±y2+1 

Since u=ex0 is positive, we must take the positive solution u+ and must discard the negative solution u−. Therefore,

ex≡u=y+y2+1⇝x=lny+y2+1 

which means that

sinh−1(y)=lny+y2+1 

Similarly, one obtains

tanh−1(y)=21ln1−y1+y 

The cosh−1 is a bit more tricky.

1.5.3 Derivatives

These are obtained by going back to the definitions of the hyperbolic functions.

sinh(x)=cosh(x)cosh(x)=sinh(x)tanh(x)=1−tanh2(x)

1.5.4 Hyperbolic Identities

These also are obtained by using the definitions of cosh and sinh:

cosh2(x)−sinh2(x)sinh(2x)==1cosh(2x)=1+2sinh2(x)2sinh(x)cosh(x)