Hyperbolic and trigonometric functions

3 Hyperbolic and trigonometric functions

We have seen in subsection 1 (Key Point 8) that

$e^{i θ} = cos θ + i sin θ$

It follows from this that

$e^{- i θ} = cos (- θ) + i sin (- θ) = cos θ - i sin θ$

Now if we add these two relations together we obtain

$cos θ = \frac{e^{i θ} + e^{- i θ}}{2}$

whereas if we subtract the second from the first we have

$sin θ = \frac{e^{i θ} - e^{- i θ}}{2 i}$

These new relations are reminiscent of the hyperbolic functions introduced in HELM booklet 6. There we defined $cosh x$ and $sinh x$ in terms of the exponential function:

$cosh x = \frac{e^{x} + e^{- x}}{2} sinh x = \frac{e^{x} - e^{- x}}{2}$

In fact, if we replace $x$ by $i θ$ in these last two equations we obtain

$cosh (i θ) = \frac{e^{i θ} + e^{- i θ}}{2} \equiv cos θ and sinh (i θ) = \frac{e^{i θ} - e^{- i θ}}{2} \equiv i sin θ$

Although, by our notation, we have implied that both $x$ and $θ$ are real quantities in fact these expressions for $cosh$ and $sinh$ in terms of $cos$ and $sin$ are quite general.

Key Point 10

If $z$ is any complex number then

cosh (i z) \equiv cos z and sinh (i z) \equiv i sin z

Equivalently, replacing

z

i z

cosh z \equiv cos (i z) and i sinh z \equiv sin (i z)

Task!

Given that ${cos}^{2} z + {sin}^{2} z \equiv 1$ for all $z$ then, utilising complex numbers, obtain the equivalent identity for hyperbolic functions.

You should obtain ${cosh}^{2} z - {sinh}^{2} z \equiv 1$ since, if we replace $z$ by $i z$ in the given identity then ${cos}^{2} (i z) + {sin}^{2} (i z) \equiv 1$ . But as noted above $cos (i z) \equiv cosh z$ and $sin (i z) \equiv i sinh z$ so the result follows.

Further analysis similar to that in the above task leads to Osborne’s rule :

Key Point 11

Osborne’s Rule

Hyperbolic function identities are obtained from trigonometric function

identities by replacing $sin θ$ by $sinh θ$ and $cos θ$ by $cosh θ$ except that

every occurrence of ${sin}^{2} θ$ is replaced by $- {sinh}^{2} θ$ .

Example 7

Use Osborne’s rule to obtain the hyperbolic identity equivalent to

$1 + {tan}^{2} θ \equiv {sec}^{2} θ$ .

Solution

Here $1 + {tan}^{2} θ \equiv {sec}^{2} θ$ is equivalent to $1 + \frac{{sin}^{2} θ}{{cos}^{2} θ} \equiv \frac{1}{{cos}^{2} θ}$ . Hence if

${sin}^{2} θ \to - {sinh}^{2} θ and {cos}^{2} θ \to {cosh}^{2} θ$

then we obtain

$1 - \frac{{sinh}^{2} θ}{{cosh}^{2} θ} \equiv \frac{1}{{cosh}^{2} θ} or, equivalently, 1 - {tanh}^{2} θ \equiv {sech}^{2} θ$