Title: Default for 6.2 The Hyperbolic Functions
Course: HELM Workbooks 2023-2024

Even and Odd Functions

Constructing even and odd functions

A given function \(f(x)\) can always be split into two parts, one of which is even and one of which is odd. To do this write \(f(x)\) as \(\dfrac{1}{2}[f(x)+f(x)]\) and then simply add and subtract \(\dfrac{1}{2}f(-x)\) to this to give \[\qquad f(x)=\dfrac{1}{2}[f(x)+f(-x)]+\dfrac{1}{2}[f(x)-f(-x)].\] The term \(\dfrac{1}{2}[f(x)+f(-x)]\) is even because when \(x\) is replaced by \(-x\) we have \(\dfrac{1}{2}[f(-x)+f(x)]\) which is the same as the original. However, the term \(\dfrac{1}{2}[f(x)-f(-x)]\) is odd since, on replacing \(x\) by \(-x\) we have \(\dfrac{1}{2}[f(-x)-f(x)]=-\dfrac{1}{2}[f(x)-f(-x)]\) which is the negative of the original.

Example

Separate \(x^3+2^x\) into odd and even parts.

Solution

\[\begin{align*}f(x) &= x^3+2^x,\\ f(-x)&=-x^3+2^{-x}. \end{align*}\]

Even part: \[\dfrac{1}{2} (f(x) + f(-x)) = \dfrac{1}{2}(x^3+2^x-x^3+2^{-x}) = \dfrac{1}{2}(2^x+2^{-x}).\]

Odd part: \[\dfrac{1}{2} (f(x) - f(-x)) = \dfrac{1}{2}(x^3+2^x+x^3-2^{-x}) = \dfrac{1}{2}(2x^3+2^x-2^{-x}).\]

Exercise

Separate the function \(f(x) = [value]\) into odd and even parts.

1. First, find \(f(-x)\).
\(f(-x) = \)
2. Next, find the even part of \(f(x)\):
3. Finally, find the odd part of \(f(x)\):
Feedback: We have the the function \(f(x) = [value].\) \[ f(-x) = [value] \] The even part is given by \[ \frac{1}{2}[f(x) + f(-x)] = [value] = [value]. \] The odd part is given by \[ \frac{1}{2}[f(x) - f(-x)] = [value] = [value]. \]

The odd and even parts of the exponential function

Using the approach outlined above we see that the even part of \(e^x\) is \(\dfrac{1}{2}(e^x+e^{-x})\) and the odd part of \(e^x\) is \(\dfrac{1}{2}(e^x-e^{-x})\) We give these new functions special names: \(\cosh x\) (pronounced ‘cosh’ \(x\)) and \(\sinh x\) (pronounced ‘shine’ \(x\)).

Key Point: Hyperbolic functions

\[\cosh x \equiv \dfrac{1}{2}(e^x+e^{-x})\]

\[\sinh x \equiv \dfrac{1}{2}(e^x-e^{-x})\]

These two functions, when added and subtracted, give \[\begin{align*} \cosh x+\sinh x &\equiv e^x, \\ \cosh x-\sinh x &\equiv e^{-x}.\end{align*}\] The graphs of \(\cosh x\) and \(\sinh x\) are shown in the below figure.

[Image]

Figure: \(\sinh x\) and \(\cosh x\)

Note that \(\cosh x>0\) for all values of \(x\) and that \(\sinh x\) is zero only when \(x=0\).

Hyperbolic Identities

The hyperbolic functions \(\cosh x\) and \(\sinh x\) satisfy similar (but not exactly equivalent) identities to those satisfied by \(\cos x\) and \(\sin x\). We note first some basic notation similar to that employed with trigonometric functions: \(\cosh^nx\) means \((\cosh x)^n\), \(\sinh^nx\) means \((\sinh x)^n\) for \(n\ne -1\).

In the special case that \(n=-1\) we do not use \(\cosh^{-1}x\) and \(\sinh^{-1}x\) to mean the reciprocals of the expressions. The notation \(\cosh^{-1}x\) and \(\sinh^{-1}x\) is reserved for the inverse functions of \(\cosh x\) and \(\sinh x\) respectively.

Exercise

We derive a fundamental identity relating hyperbolic functions \(\cosh^2x-\sinh^2x \equiv \ ?\)

1. First, express \(\cosh^2x\) in terms of the exponential functions. Expand your answer.
\(\cosh^2x \equiv \)
2. Similarly, express \(\sinh^2x\) in terms of the exponential functions. Expand your answer.
\(\sinh^2x \equiv \)
3. Finally determine \(\cosh^2x-\sinh^2x\) using the results from (a) and (b):
\(\cosh^2x-\sinh^2x \equiv \)

Alternatively, we can prove the above identity by using the relations \(e^x \equiv \cosh x + \sinh x\) and \(e^{-x} \equiv \cosh x - \sinh x .\)

See solution/feedback if you cannot work this out on your own.
Feedback: Here is a step by step calculation. \[ \begin{align*} \cosh^2x & \equiv \left( [value] \right)^2 \equiv \frac{1}{4}(e^x+e^{-x})^2 \\ & \equiv \frac{1}{4}[(e^x)^2+2e^xe^{-x}+(e^{-x})^2] \equiv \frac{1}{4}[e^{2x}+2e^{x-x}+e^{-2x}] \\ & \equiv [value] .\end{align*} \] \[ \begin{align*} \sinh^2x\ & \equiv \left( [value] \right)^2 \equiv \frac{1}{4}(e^x-e^{-x})^2 \\ & \equiv \frac{1}{4}[(e^x)^2-2e^xe^{-x}+(e^{-x})^2] \equiv \frac{1}{4}[e^{2x}-2e^{x-x}+e^{-2x}] \\ & \equiv [value]. \end{align*} \] Alternatively, using \(e^x \equiv \cosh x + \sinh x\) and \(e^{-x} \equiv \cosh x - \sinh x \), we obtain \[1 \equiv e^x \cdot e^{-x} \equiv (\cosh x + \sinh x)(\cosh x - \sinh x) \equiv \cosh^2x - \sinh^2x \] as desired.

Exercise

In this exercise, we show that \(\cosh(x+y) \equiv \cosh x \cosh y + \sinh x \sinh y\).

1. First, express \(\cosh x \cosh y \) in terms of exponentials.
\(\cosh x \cosh y \equiv  \)
2. Next, express \(\sinh x \sinh y \) in terms of exponentials.
\(\sinh x \sinh y  \equiv \)
3. Now, express \(\cosh x \cosh y + \sinh x \sinh y\) in terms of exponentials using (a) and (b),
\(\cosh x \cosh y + \sinh x \sinh y \equiv \)
and realize that this equals \(\cosh(x+y)\) by definition.
Feedback: (a) By definition of hyperbolic functions, we have \( \cosh x \cosh y \equiv \left( [value] \right) \cdot \left( [value] \right) \equiv [value] \). (b) Similarly, we have \( \sinh x \sinh y \equiv \left( [value] \right) \cdot \left( [value] \right) \equiv [value] \). (c) From (a) and (b), we deduce \[ \begin{align*} & \cosh x \cosh y + \sinh x \sinh y \\ & = [value] + [value] \\ & = [value] \\ & = \cosh (x+y). \end{align*}\]

Related Hyperbolic Functions

Given the trigonometric functions \(\cos x,\sin x\) related functions can be defined; \(\tan x, \sec x, \rm cosec\) \(x\) through the relations: \(\tan x \equiv\dfrac{\sin x}{\cos x}\qquad \sec x\equiv\dfrac{1}{\cos x}\qquad{\rm cosec}\ x\equiv\dfrac{1}{\sin x}\qquad\cot x\equiv\dfrac{\cos x}{\sin x}\).

In an analogous way, given \(\cosh x\) and \(\sinh x\) we can introduce hyperbolic functions. These functions are defined in the following Key Point.

Key Point: Further hyperbolic functions

\(\tanh x \equiv \dfrac{\sinh x}{\cosh x}\)

\(\rm sech\) \(x \equiv \dfrac{1}{\cosh x}\)

\(\rm cosech\) \(x \equiv \dfrac{1}{\sinh x}\)

\(\coth x \equiv \dfrac{\cosh x}{\sinh x}\)

Q: Exercise

Which of the following identities are correct?
- \( 1 - \tanh^2x \equiv \sec{\rm h}^2x \)
- \( 1 - \tanh^2x \equiv \sinh^2x \)
- \( 1 - \tanh^2x \equiv \cosh^2x \)
- \( 1 - \tanh^2x \equiv \dfrac{1}{\sinh^2x} \)
- \( 1 - \tanh^2x \equiv \dfrac{1}{\cosh^2x} \)
- \( 1 - \tanh^2x \equiv \dfrac{\cosh^2x - \sinh^2x}{\cosh^2x}\)
- \(1 - \tanh^2x \equiv \dfrac{\sinh^2x - \cosh^2x}{\sinh^2x}\)
A: \( 1 - \tanh^2x \equiv \sec{\rm h}^2x \); \( 1 - \tanh^2x \equiv \dfrac{1}{\cosh^2x} \); \( 1 - \tanh^2x \equiv \dfrac{\cosh^2x - \sinh^2x}{\cosh^2x}\)
Feedback: Note the following \( 1 - \tanh^2 x = 1 - \dfrac{\sinh^2 x}{\cosh^2 x} = \dfrac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x} = \) \( \rm{sech}^2 \) \(x\). To see that each of the other identities is incorrect, specify a particular value for \(x\) for which the identity is wrong. Also note that they are inconsistent with the above correct identities.

Q: Exercise

Which of the following identities are correct?
- \( \sinh 2x \equiv 2 \sinh x \cosh x \)
- \( \cosh 2x \equiv \cosh^2x + \sinh^2x \)
- \( \cosh 2x \equiv 2 \cosh^2x - 1 \)
- \( \cosh 2x \equiv 1 + 2 \sinh^2 x \)
- \( \cosh^2x + \sinh^2x \equiv 1 \)
- \( \cosh 2x \equiv - 2 \sinh x \cosh x \)
- \( \sinh 2x \equiv \sinh^2x - \cosh^2x \)
- \( \sinh 2x \equiv 2 \sinh^2x - 1 \)
- \( \sinh 2x \equiv 1 + 2 \cosh^2 x \)
A: \( \sinh 2x \equiv 2 \sinh x \cosh x \); \( \cosh 2x \equiv \cosh^2x + \sinh^2x \); \( \cosh 2x \equiv 2 \cosh^2x - 1 \); \( \cosh 2x \equiv 1 + 2 \sinh^2 x \)
Feedback: Note the following. \( \sinh(2x) = \sinh (x+x) \), so using the formula for \( \sinh(x+y) \) with \( y = x \) we obtain \( \sinh(2x) = 2 \sinh x \cosh x \). \( \cosh(2x) = \cosh (x+x) \), so using the formula for \( \cosh(x+y) \) with \( y = x \) we obtain \( \cosh(2x) = \cosh^2 x + \sinh^ 2 x \). Recall that \( \cosh^2 x - \sinh^2 x = 1 \) (previous key point) which implies \( \sinh^2 x = \cosh^2 x - 1 \), so combining with the formula for \(\cosh^2 x\) just obtained above we get \[ \cosh(2x) = \cosh^2 x + \sinh^ 2 x = \cosh^2 x + ( \cosh^2 x - 1) = 2 \cosh^2 x - 1.\] Also, \( \cosh^2 x - \sinh^2 x = 1 \) (previous key point) implies \( \cosh^2 x = 1 + \sinh^2 x\), so combining with the formula for \(\cosh^2 x\) just obtained above we get \[ \cosh(2x) = \cosh^2 x + \sinh^ 2 x = (1+ \sinh^2 x) + \sinh^2 x = 1 + 2 \sinh^2 x.\] To see that each of the other identities is incorrect, specify a particular value for \(x\) for which the identity is wrong. Also note that they are inconsistent with the correct identities that are proved above.

Exercise

In this exercise, we show that \(\sinh(x+y) \equiv \sinh x \cosh y + \cosh x \sinh y\).

1. First, express \(\sinh x \cosh y \) in terms of exponentials.
\(\sinh x \cosh y  \equiv \)
2. Next, express \(\cosh x \sinh y \) in terms of exponentials.
\(\cosh x \sinh y  \equiv \)
3. Now, express \(\sinh x \cosh y + \cosh x \sinh y\) in terms of exponentials using (a) and (b),
\(\sinh x \cosh y + \cosh x \sinh y \equiv \)
and realize that this equals \(\sinh(x+y)\) by definition.
Feedback: (a) By definition of hyperbolic functions, we have \( \sinh x \cosh y \equiv \left( [value] \right) \cdot \left( [value] \right) \equiv [value] \). (b) Similarly, we have \( \cosh x \sinh y \equiv \left( [value] \right) \cdot \left( [value] \right) \equiv [value] \). (c) From (a) and (b), we deduce \[ \begin{align*} & \sinh x \cosh y + \cosh x \sinh y \\ & = [value] + [value] \\ & = [value] \\ & = \sinh (x+y). \end{align*}\]

Key Point

The fundamental identity relating hyperbolic functions is:\[\cosh^2x - \sinh^2x \equiv 1.\] This is the hyperbolic function equivalence of the trigonometric identity: \(\cos^2x + \sin^2x \equiv 1\).

Key Point: More hyperbolic identities

\[\begin{align*} \cosh(x+y) &\equiv \cosh x \cosh y+\sinh x \sinh y \\ \sinh(x+y) &\equiv \sinh x \cosh y+\cosh x\sinh y \\ \sinh2x &\equiv 2\sinh x \cosh x \\ \cosh2x &\equiv \cosh^2x+\sinh^2x \equiv 2\cosh^2x-1 \equiv 1+2\sinh^2x \end{align*} \]

Exercise

Express the following in terms of \(e^{[value][value]}\) and \(e^{-[value][value]}\).

\(  [value] \sinh([value][value]) + [value] \cosh([value][value]) = \)
Feedback: \[\begin{align*} & [value] \sinh([value][value]) + [value] \cosh([value][value]) \\ & = [value] \cdot \frac{1}{2} (e^{[value][value]} - e^{-[value][value]}) + [value] \cdot \frac{1}{2} (e^{[value][value]} + e^{-[value][value]}) \\ & = [value] \end{align*}\]

Exercise

Express the following in terms of \(e^{[value][value]}\) and \(e^{-[value][value]}\).

\(  [value] \sinh([value][value]) - [value] \cosh([value][value]) = \)
Feedback: \[\begin{align*} & [value] \sinh([value][value]) - [value] \cosh([value][value]) \\ & = [value] \cdot \frac{1}{2} (e^{[value][value]} - e^{-[value][value]}) - [value] \cdot \frac{1}{2} (e^{[value][value]} + e^{-[value][value]}) \\ & = [value] \end{align*}\]

Exercise

Express the following in terms of \(\sinh([value][value])\) and \(\cosh([value][value])\).

\(  [value] e^{[value][value]} - [value] e^{-[value][value]} = \)
Feedback: Using \(e^{[value][value]} = \cosh [value][value] + \sinh [value][value] \) and \(e^{-[value][value]} = \cosh [value][value] - \sinh [value][value] \), we get \[ \begin{align*} & [value] e^{[value][value]} - [value] e^{-[value][value]} \\ & = [value]( \cosh [value][value] + \sinh [value][value] ) - [value] (\cosh [value][value] - \sinh [value][value]) \ \\ & = [value] \end{align*}\]

Exercise

Express the following in terms of \(\sinh([value][value])\) and \(\cosh(-[value][value])\).

\(  [value] e^{-[value][value]} - [value] e^{[value][value]} = \)
Feedback: Using \(e^{[value][value]} = \cosh [value][value] + \sinh [value][value] \) and \(e^{-[value][value]} = \cosh [value][value] - \sinh [value][value]\), we get \[ \begin{align*} & [value] e^{-[value][value]} - [value] e^{[value][value]} \\ & = [value]( \cosh [value][value] - \sinh [value][value] ) - [value] (\cosh [value][value] + \sinh [value][value]) \ \\ & = [value] \end{align*}\]

Exercise

Express the following in terms of \(\coth([value][value])\).

\(  \dfrac{[value] e^{[value][value]}} {[value] e^{[value][value]} - [value] e^{-[value][value]}} = \)
Feedback: It will be worth keeping in mind that \(\coth(z)=\dfrac{\cosh(z)}{\sinh(z)}\). We start by rewriting the expression in terms of \(\sinh([value][value])\) and \(\cosh([value][value])\) as at least these are more like the expression we are targeting. For the numerator we use \[e^z=\cosh(z)+\sinh(z)\] while for denominator we can notice that this is already quite close to \(\sinh([value][value])\) indeed \[ \sinh([value][value])=\frac{e^{[value][value]}-e^{-[value][value]}}{2}.\] Note that instead of spotting this we could just write each \(e^z=\cosh(z)+\sinh(z)\), there would be some cancelation after using the even / odd properties of \(\cosh(z)\) and \(\sinh(z)\). The result is of course the same. It's good exercise to confirm. In any case, we can bring this together and simplify to see \[\dfrac{[value] e^{[value][value]}} {[value] e^{[value][value]} - [value] e^{-[value][value]}} = \dfrac{[value](\cosh([value][value]) + \sinh([value][value]))}{[value] \sinh([value][value])} = \dfrac{[value]}{[value]} \cdot \frac{\cosh([value][value])}{\sinh([value][value])} + \dfrac{[value]}{[value]} = [value]. \]

Exercise

Using only the \(\cosh\) and \(\sinh\) keys on your calculator (or \(e^x\) key) find the values of the following. Round up and enter 4 decimal places. If the answer has less than 4 decimal places, then enter zero(s) at the end.

1. \( \tanh [value] =\)
2. \( \rm{cosech} \ [value] =\)
3. \( \rm{sech} \ [value] =\)
Feedback: (a) Using \( \cosh\) and \(\sinh \) keys, \( \tanh [value] = \dfrac{\sinh [value]}{\cosh [value]} = [value] \) Using \(e^x\) key, \( \tanh [value] = \dfrac{e^{[value]} - e^{-[value]}}{e^{[value]} + e^{-[value]}} = [value]\) (b) Using \(\sinh \) key, \( \rm{cosech} \ [value] = \dfrac{1}{\sinh [value]} = [value]\) Using \(e^x\) key, \( \rm{cosech} \ [value] = \dfrac{1}{e^{[value]} - e^{-[value]}} = [value] \) (c) Using \( \cosh\) key, \( \rm{sech} \ [value] = \dfrac{1}{\cosh [value]} = [value]\) Using \(e^x\) key, \( \rm{sech} \ [value] = \dfrac{1}{e^{[value]} + e^{-[value]}} = [value] \)

Exercise

Given \(\tanh[value] = [value] \). Compute \( \rm{sech}\) \( [value] \). Round up and enter 4 decimal places. If the answer has less than 4 decimal places, then enter zero(s) at the end.

\( \rm{sech} \) \([value] = \)
Feedback: Use the identity \( \rm{sech}^2\)\([value] = 1 - \tanh^2[value]\) to calculate \(\rm{sech}^2\)\([value]\), then take the (positive) square root to obtain \(\rm{sech}\) \([value] \), noting that \(\rm{sech}\) \( [value] = \frac{1}{\cosh[value]} >0\) for all values of \([value]\). \(\rm{sech}\) \([value] = \sqrt{1 - \tanh^2[value]} = \sqrt{1 - ([value])^2} =[value]\) to four significant figures (or four decimal places in this case).

Exercise

Let \( \sinh x = [value] \), and \( \sinh y = [value] \), calculate \(\cosh(x+y)\).

Note: Write down a formula for \( \cosh(x+y) \) in terms of \( \sinh x, \sinh y \), and only sub in numbers and use a calculator at the final step.

Round up your answer and enter \(4\) decimal places. If the answer has fewer than \(4\) decimal places, then enter zero(s) at the end.

\( \cosh(x+y) = \)
Feedback: Using the fundamental identity \( \cosh^2x - \sinh^2x = 1\), we get \(\cosh^2 x = 1 + \sinh^2 x\). Similarly, \(\cosh^2 y = 1 + \sinh^2 y\). Next, take the (positive) square root to obtain \( \cosh x\), noting that \( \cosh x > 0 \) for all value of \(x\). Similar for \( \cosh y \). Then use the formula \( \cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y \), we obtain \[\begin{align*} \cosh(x+y) & = \cosh x \cosh y + \sinh x \sinh y \\ & = \sqrt{1 + \sinh^2 x} \cdot \sqrt{1 + \sinh^2 y} + \sinh x \sinh y \\ & = \sqrt{1 + ([value])^2} \cdot \sqrt{1 +([value])^2} + ([value]) \cdot ([value])\\ & = [value] \text{ after rounding up.} \end{align*}\]

Exercise

Let \( \sinh x = [value] \), and \( \sinh y = [value] \), calculate \(\sinh(x+y)\).

Note: Write down a formula for \( \sinh(x+y) \) in terms of \( \sinh x, \sinh y \), and only sub in numbers and use a calculator at the final step. Round up your answer and enter 4 decimal places. If the answer has less than 4 decimal places, then enter zero(s) at the end.

\( \sinh(x+y) = \)
Feedback: Using the fundamental identity \( \cosh^2x - \sinh^2x = 1\), we get \(\cosh^2 x = 1 + \sinh^2 x\). Similarly, \(\cosh^2 y = 1 + \sinh^2 y\). Next, take the (positive) square root to obtain \( \cosh x\), noting that \( \cosh x > 0 \) for all value of \(x\). Similar for \( \cosh y \). Then use the formula \( \sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y \), we obtain \[\begin{align*} \sinh(x+y) & = \sinh x \cosh y + \cosh x \sinh y \\ & = \sinh x \cdot \sqrt{1 + \sinh^2 y} + \sqrt{1 + \sinh^2 x} \cdot \sinh y \\ & = ([value]) \cdot \sqrt{1 + ([value])^2} + \sqrt{1 +([value])^2} \cdot ([value]) \\ & = [value] \text{ after rounding up.} \end{align*}\]

Key Point:

The fundamental identity relating hyperbolic functions is:

\(\rm cosh^2 x - \sinh^2 x \equiv 1\)

This is the hyperbolic function equivalent of the trigonometric identity: \( \cos^2x + \sin^2 x \equiv 1 \)