Learning Lab

Example 1

What is the exact value of \(\sinh\left(0\right)?\)

Solution:

By definition, \[\begin{align*} \sinh\left(0\right) & =\frac{e^{0}-e^{-0}}{2}\\ & =\frac{1-1}{2}\\ & =0. \end{align*}\]

Hence the value of \(\sinh\left(0\right)=0.\)

Example 2

What is the exact value of \(\cosh\left(\ln\left(2\right)\right)\)?

Solution:

By definition, \[\begin{align*} \cosh\left(\ln\left(2\right)\right) & =\frac{1}{2}\left(e^{\ln\left(2\right)}+e^{-\ln\left(2\right)}\right)\\ & =\frac{1}{2}\left(e^{\ln\left(2\right)}+e^{\ln\left(2^{-1}\right)}\right)\\ & =\frac{1}{2}\left(e^{\ln\left(2\right)}+e^{\ln\left(\frac{1}{2}\right)}\right)\\ & =\frac{1}{2}\left(2+\frac{1}{2}\right)\\ & =\frac{1}{2}\cdot\frac{5}{2}\\ & =\frac{5}{4}. \end{align*}\]

Hence \(\cosh\left(\ln\left(2\right)\right)=5/4.\)

Example 3

Solve for \(x,\) \(\sinh\left(x\right)=3/4\).

Solution:

Using the definition of \(\sinh,\) \[\begin{align*} \frac{1}{2}\left(e^{x}-e^{-x}\right) & =\frac{3}{4}\\ e^{x}-e^{-x} & =\frac{3}{2}. \end{align*}\] Multiplying both sides by \(2e^{x}\) and rearranging, \[\begin{align*} 2e^{2x}-2 & =3e^{x}\\ 2e^{2x}-3e^{x}-2 & =0\\ 2\left(e^{x}\right)^{2}-3e^{x}-2 & =0, \end{align*}\] which is a quadratic in \(e^{x}.\) Using the quadratic formula11 For a quadratic \[\begin{align*} ax^{2}+bx+c & =0 \end{align*}\] where \(a,\) \(b\) and \(c\) are constants, \[\begin{align*} x & =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. \end{align*}\] \[\begin{align*} e^{x} & =\frac{3\pm\sqrt{\left(-3\right)^{2}-4\left(2\right)\left(-2\right)}}{2\left(2\right)}\\ & =\frac{3\pm\sqrt{9+16}}{4}\\ & =\frac{3\pm5}{4}. \end{align*}\] Since \(e^{x}>0\) for all \(x,\) we ignore the negative sign and \[\begin{align*} e^{x} & =\frac{3+5}{4}\\ & =2. \end{align*}\] Taking logs of both sides gives \(x=\ln\left(2\right).\) Hence, the required value of \(x=\ln\left(2\right).\)

Example 4

If \(\sinh\left(x\right)=3/5,\) what is the value of \(\cosh\left(x\right)?\)

Solution:

Using the identity \[\begin{align*} \cosh^{2}\left(x\right)-\sinh^{2}\left(x\right) & =1 \end{align*}\] we have \[\begin{align*} \cosh^{2}\left(x\right) & =1+\sinh^{2}\left(x\right)\\ & =1+\left(\frac{3}{5}\right)^{2}\\ & =\frac{34}{25}\\ \cosh\left(x\right) & =\pm\sqrt{\frac{34}{25}}\\ & =\frac{\sqrt{34}}{5}. \end{align*}\] Note that we reject \(-\sqrt{34}/5\) as the range of \(\cosh\) is \(\left[\left.1,\infty\right)\right..\)

Example 5

If \(\cosh\left(x\right)=5/3,\) what is the value of \(\sinh\left(x\right)\)?

Solution:

Using the identity \[\begin{align*} \cosh^{2}\left(x\right)-\sinh^{2}\left(x\right) & =1 \end{align*}\] we have \[\begin{align*} \sinh^{2}\left(x\right) & =\cosh^{2}\left(x\right)-1\\ & =\left(\frac{5}{3}\right)^{2}-1\\ & =\frac{25}{9}-1\\ & =\frac{16}{9}\\ \sinh\left(x\right) & =\pm\sqrt{\frac{16}{9}}\\ & =\pm\frac{4}{3}. \end{align*}\] Note that there are two solutions in this case because the range of \(\sinh\) is \(\left(-\infty,\infty\right)\) and \(\cosh\left(x\right)\) is not one to one.

Example 6

If \(\sinh\left(x\right)=3/4,\) what is the exact value of \(x?\)

Solution:

From the definition of \(\sinh\left(x\right)\) \[\begin{align*} \frac{3}{4} & =\frac{1}{2}\left(e^{x}-e^{-x}\right). \end{align*}\] Multiplying both sides by 4 and rearranging, \[\begin{align*} 2e^{x}-3-2e^{-x} & =0. \end{align*}\] Multiplying both sides22 No problems arise from this as \(e^{x}>0\) for all \(x.\) by \(e^{x},\) \[\begin{align*} 2e^{2x}-3e^{x}-2 & =0. \end{align*}\] This is a quadratic in \(e^{x}\) and may be factorised to give \[\begin{align*} \left(2e^{x}+1\right)\left(e^{x}-2\right) & =0. \end{align*}\] The first bracket is never zero as \(e^{x}>0\) for all \(x.\) Hence \[\begin{align*} e^{x}-2 & =0\\ e^{x} & =2\\ x & =\ln\left(2\right). \end{align*}\]

The value of \(x\) is \(\ln\left(2\right).\)

Example 7

Show that \(\sinh\left(2x\right)=2\sinh\left(x\right)\cosh\left(x\right).\)

Solution:

In this solution, we use the difference of two squares formula33 Difference of two squares formula is \[\begin{align*} a^{2}-b^{2} & =\left(a-b\right)\left(a+b\right). \end{align*}\]. Using the definition of \(\sinh,\) the left hand side may be written as \[\begin{align*} \sinh\left(2x\right) & =\frac{1}{2}\left(e^{2x}-e^{-2x}\right)\\ & =\frac{1}{2}\left(\left(e^{x}\right)^{2}-\left(e^{-x}\right)^{2}\right)\\ & =\frac{1}{2}\left(e^{x}-e^{-x}\right)\left(e^{x}+e^{-x}\right)\\ & =\sinh\left(x\right)\left(e^{x}+e^{-x}\right)\\ & =2\sinh\left(x\right)\frac{\left(e^{x}+e^{-x}\right)}{2}\\ & =2\sinh\left(x\right)\cosh\left(x\right) \end{align*}\] as required.

Example 8

Given that \(\tanh\left(x\right)=4/5,\) find the value of all the other hyperbolic functions: \(\sinh,\) \(\cosh,\) \(\ \text{sech, coshec and coth.}\)

Solution:

Using the identity \(1-\tanh^{2}\left(x\right)=\text{sech$^{2}\left(x\right)$ }\), \[\begin{align*} \text{sech$\left(x\right)$ } & =\sqrt{1-\left(\frac{4}{5}\right)^{2}}\\ & =\sqrt{\frac{9}{25}}\\ & =\frac{3}{5} \end{align*}\] and \[\begin{align*} \cosh\left(x\right) & =\frac{5}{3}. \end{align*}\]

Using the identity \(\cosh^{2}\left(x\right)-\sinh^{2}\left(x\right)=1\), \[\begin{align*} \sinh\left(x\right) & =\sqrt{\left(\frac{5}{3}\right)^{2}-1}\\ & =\sqrt{\frac{16}{9}}\\ & =\frac{4}{3} \end{align*}\] and \[\begin{align*} \text{cosech$\left(x\right)$ } & =\frac{3}{4}. \end{align*}\]

Note that we could also use \[\begin{align*} \tanh\left(x\right) & =\frac{\sinh\left(x\right)}{\cosh\left(x\right)}\\ \sinh\left(x\right) & =\tanh\left(x\right)\cosh\left(x\right)\\ & =\frac{4}{5}\times\frac{5}{3}\\ & =\frac{4}{3} \end{align*}\] to get \(\sinh\left(x\right)\) and hence cosech\(\left(x\right).\)

Finally, given that \(\tanh\left(x\right)=4/5\), \[\begin{align*} \text{coth$\left(x\right)$ } & =\frac{5}{4}. \end{align*}\]

Answers

1. 1. \(x+\frac{1}{x}\qquad\)b) \(\frac{1}{2}\left(x^{2}-\frac{1}{x^{2}}\right)\)
2. \(\tanh\left(x\right)=\frac{3}{5}\).