Inverse Trigonometric and Hyperbolic Functions

Section 5.5 Inverse Trigonometric and Hyperbolic Functions

We expressed trigonometric and hyperbolic functions in Section 5.4 in terms of the exponential function. In this section we look at their inverses. When we solve equations such as \(w=\sin z\) for \(z\text{,}\) we obtain formulas that involve the logarithm. Because trigonometric and hyperbolic functions are all periodic, they are many-to-one; hence their inverses are necessarily multivalued. The formulas for the inverse trigonometric functions are

\begin{align} \arcsin z \amp = -i\log \left[ iz+(1-z^2 )^{\frac{1}{2}}\right],\tag{5.5.1}\\ \arccos z \amp = -i\log \left[ z+i(1-z^2 )^{\frac{1}{2}}\right], \text{ and }\notag\\ \arctan z \amp = \frac{i}{2}\log\left(\frac{i+z}{i-z}\right)\text{.}\notag \end{align}

We can find the derivatives of any branch of these functions by using the chain rule:

\begin{align} \frac{d}{dz}\arcsin z \amp = \frac{1}{(1-z^2 )^{\frac{1}{2}}},\tag{5.5.2}\\ \frac{d}{dz}\arccos z \amp = \frac{-1}{(1-z^2 )^{\frac{1}{2}}}, \text{ and }\notag\\ \frac{d}{dz}\arctan z \amp = \frac{1}{1+z^2 }\text{.}\notag \end{align}

We derive Equations (5.5.1) and (5.5.2) and leave the others as exercises. If we take a particular branch of the multivalued function, \(w=\arcsin z\text{,}\) we have

\begin{equation*} z=\sin w=\frac{1}{2i}(e^{iw}-e^{-iw})\text{,} \end{equation*}

which we can also write as

\begin{equation*} e^{iw}-2iz-e^{-iw}=0\text{.} \end{equation*}

Multiplying both sides of this equation by \(e^{iw}\) gives \((e^{iw})^2 -2ize^{iw}-1=0\text{,}\) which is a quadratic equation in terms of \(e^{iw}\text{.}\) Using the quadratic equation to solve for \(e^{iw}\text{,}\) we obtain

\begin{equation*} e^{iw}=\frac{2iz+(4-4z^2 )^{\frac{1}{2}}}{2}=iz+(1-z^2 )^{\frac{1}{2}}\text{,} \end{equation*}

where the square root is a multivalued function. Taking the logarithm of both sides of this last equation leads to the desired result:

\begin{equation*} w=\arcsin z=-i\log \left[ iz+(1-z^2 )^{\frac{1}{2}}\right]\text{.} \end{equation*}

where the multivalued logarithm is used. To construct a specific branch of \(\arcsin z\text{,}\) we must first select a branch of the square root and then select a branch of the logarithm.

We get the derivative of \(w=\arcsin z\) by starting with the equation \(\sin w=z\) and differentiating both sides, using the chain rule:

\begin{align*} \frac{d}{dz}\sin w \amp = \frac{d}{dz}z,\\ \frac{d}{dw}\sin w\frac{dw}{dz} \amp = 1,\\ \frac{dw}{dz} \amp = \frac{1}{\cos w}\text{.} \end{align*}

When the principal value is used, \(w=\arcsin z=-i\mathrm{Log}\left[ iz+(1-z^2 )^{\frac{1}{2}}\right]\) maps the upper half-plane \(\{z:\mathrm{Im}(z) >0\}\) onto a portion of the upper half-plane \(\{w:\mathrm{Im}(w) >0\}\) that lies in the vertical strip \(\{w:\frac{-\pi}{2}\lt \mathrm{Re}(w) \lt \frac{\pi}{2}\}\text{.}\) The image of a rectangular grid in the \(z\) plane is a “spider web” in the \(w\) plane, as Figure 5.5.1 shows.

Figure 5.5.1. A rectangular grid is mapped onto a spider web by \(w=\arcsin z\)

Example 5.5.2.

The values of \(\arcsin \sqrt{2}\) are given by

\begin{equation} \arcsin \sqrt{2} = -i\log \left[ i\sqrt{2}+ \left(1-\left(\sqrt{2}\right)^2\right)^{\!{\frac{1}{2}}}\right] =-i\log(i\sqrt{2}\pm i)\text{.}\tag{5.5.3} \end{equation}

Using straightforward techniques, we simplify this equation and obtain

\begin{align*} \arcsin \sqrt{2} \amp = -i\log \left[(\sqrt{2}\pm 1) i\right]\\ \amp = i\left[ \ln(\sqrt{2}\pm 1) +i\left(\frac{\pi}{2}+2n\pi\right)\right]\\ \amp = \frac{\pi}{2}+2n\pi -i\ln(\sqrt{2}\pm 1), \text{ where } n \text{ is an integer. } \end{align*}

We observe that

\begin{equation*} \ln(\sqrt{2}-1) =\ln \frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}=\ln \frac{1}{\sqrt{2}+1}=-\ln(\sqrt{2}+1) \end{equation*}

and then write

\begin{equation*} \arcsin \sqrt{2}=\frac{\pi}{2}+2n\pi \pm i\ln(\sqrt{2}+1), \text{ where } n \text{ is an integer } \end{equation*}

Example 5.5.3.

Suppose that we make specific choices in Equation (5.5.3) by selecting \(+i\) as the value of the square root \([1-(\sqrt{2})^2]^{\frac{1}{2}}\) and using the principal value of the logarithm. With \(f(z) =\mathrm{Arcsin}\,z\text{,}\) The result is

\begin{equation*} f(\sqrt{2}) =\mathrm{Arcsin}\sqrt{2}=-i\mathrm{Log}(i\sqrt{2}+i) =\frac{\pi}{2}-i\ln(\sqrt{2}+1)\text{,} \end{equation*}

and the corresponding value of the derivative is given by

\begin{equation*} f\,'(\sqrt{2}) =\frac{1}{\left[ 1-\left(\sqrt{2}\right)^2 \right]^{\frac{1}{2}}}=\frac{1}{i}=-i \end{equation*}

The inverse hyperbolic functions are

\begin{align} \text{ arcsinh\ } z \amp = \log \left[ z+(z^2 +1)^{\frac{1}{2}} \right],\notag\\ \text{ arccosh\ } z \amp = \log \left[ z+(z^2 -1)^{\frac{1}{2}} \right] \text{ , and }\notag\\ \text{ arctanh\ } z \amp = \frac{1}{2}\log\left(\frac{1+z}{1-z}\right)\text{.}\tag{5.5.4} \end{align}

Their derivatives are

\begin{align*} \frac{d}{dz}\text{ arcsinh\ } z \amp = \frac{1}{(z^2 +1)^{\frac{1}{2}}},\\ \frac{d}{dz}\text{ arccosh\ } z \amp = \frac{1}{(z^2 -1)^{\frac{1}{2}}}, \text{ and }\\ \frac{d}{dz}\text{ arctanh\ } z \amp = \frac{1}{1-z^2 }\text{.} \end{align*}

To establish Identity (5.5.4), we start with \(w=\) arctanh\(z\) and obtain

\begin{equation*} z=\tanh w=\frac{e^{w}-e^{-w}}{e^{w}+e^{-w}}=\frac{e^{2w}-1}{e^{2w}+1}\text{.} \end{equation*}

which we solve for \(e^{2w}\text{,}\) getting \(e^{2w}=\frac{1+z}{1-z}\text{.}\) Taking the logarithms of both sides gives

\begin{equation*} w=\text{ arctanh\ } z=\frac{1}{2}\log\left(\frac{1+z}{1-z}\right)\text{,} \end{equation*}

which is what we wanted to show.

Example 5.5.4.

Calculation reveals that

\begin{align*} \text{ arctanh } (1+2i) \amp = \frac{1}{2}\log \frac{1+1+2i}{1-1-2i}=\frac{1}{2}\log(-1+i)\\ \amp = \frac{1}{4}\ln 2+i\left(\frac{3}{8}+n\right)\pi\text{,} \end{align*}

where \(n\) is an integer.

Exercises Exercises

1.

Find all values of the following.

(a)

\(\arcsin \frac{5}{4}\text{.}\)

Solution.

\((\frac{1}{2}+2n) \pi \pm i\ln 2\text{,}\) where \(n\) is an integer.

(b)

\(\arccos \frac{5}{3}\text{.}\)

(c)

\(\arcsin 3\text{.}\)

Solution.

\((\frac{1}{2}+2n) \pi \pm i\ln (3 + 2\sqrt{2})\text{,}\) where \(n\) is an integer.

(d)

\(\arccos 3i\text{.}\)

(e)

\(\arctan 2i\text{.}\)

Solution.

\(-(\frac{1}{2}+n) \pi +i\ln \sqrt{3}\text{,}\) where \(n\) is an integer.

(f)

\(\arctan i\text{.}\)

(g)

\(\mathrm{arcsinh}\,i\text{.}\)

Solution.

\(i(\frac{1}{2}+2n) \pi\text{,}\) where \(n\) is an integer.

(h)

\(\mathrm{arcsinh}\, \frac{3}{4}\text{.}\)

(i)

\(\mathrm{arccosh}\, i\text{.}\)

Solution.

\(\ln (\sqrt{2}+1) + i(\frac{1}{2}+2n)\pi\) and \(\ln(\sqrt{2}-1) + i(-\frac{1}{2} +2n)\pi\text{,}\) where \(n\) is an integer.

(j)

\(\mathrm{arccosh}\, \frac{1}{2}\text{.}\)

(k)

\(\mathrm{arctanh}\, i\text{.}\)

Solution.

\(i(\frac{1}{4}+n) \pi\text{,}\) where \(n\) is an integer.

(l)

\(\mathrm{arctanh}\, i\sqrt{3}\text{.}\)

2.

Establish the following identities .

(a)

\(\arccos z=-i\log \left[z+i(1-z^2 )^{\frac{1}{2}}\right]\text{.}\)

(b)

\(\frac{d}{dz}\arccos z=\frac{-1}{(1-z^2 )^{\frac{1}{2}}}\text{.}\)

(c)

\(\arctan z=\frac{i}{2}\log(\frac{i+z}{i-z})\text{.}\)

(d)

\(\frac{d}{dz}\arctan z=\frac{1}{1+z^2 }\text{.}\)

(e)

\(\arcsin z+\arccos z=\frac{\pi}{2}+2n\pi\text{,}\) where \(n\) is an integer.

(f)

\(\frac{d}{dz}\mathrm{arctanh}\,z=\frac{1}{1-z^2}\text{.}\)

(g)

\(\mathrm{arcsinh}\,z=\log \left[z+(z^2 +1)^{\frac{1}{2}}\right]\text{.}\)

(h)

\(\frac{d}{dz}\mathrm{arcsinh}\,z=\frac{1}{(z^2 +1)^{1/2}}\text{.}\)

(i)

\(\mathrm{arccosh}\,z=\log\left[ z+(z^2 -1)^{\frac{1}{2}}\right]\text{.}\)

(j)

\(\frac{d}{dz}\mathrm{arccosh}\,z=\frac{1}{(z^2 -1)^{\frac{1}{2}}}\text{.}\)

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