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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 6.6 Inverse Trigonometric Functions

Subsection 6.6.1 Before Class

https://mymedia.ou.edu/media/6.6-1/1_bv2ywri0
Figure 55. Pre-Class Video 1
https://mymedia.ou.edu/media/6.6-2/1_8spfkiyw
Figure 56. Pre-Class Video 2

Subsubsection 6.6.1.1 Building Inverse Trig Functions

Question 6.6.1.
Refer back to Section 6.1-how can we determine if a function has an inverse?
Solution.
The function must be one-to-one on the interval in question.
Example 6.6.2.
Let \(f(x) = \sin x\)
  1. Graph \(\sin x\) on the interval \([-2\pi,2\pi]\)
  2. Identify an interval on which \(\sin x\) could possess an inverse
  3. Sketch the graph of the inverse function
Solution.
  1. The graph of \(\sin x\) on the interval \([-2\pi,2\pi]\)
  2. Sine achieves all its potential values on the interval \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\text{,}\) so use this as our restricted domain.
  3. The graph of \(\arcsin x\)
Inverse Sine (Arcsine).
The function \(y = \sin x\) has the inverse \(x=\inv{\sin}y\) (also written \(\arcsin y\)). The domain of \(\inv{\sin}y\) is \([-1,1]\) and the range is \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\)
Example 6.6.3.
Evaluate \(\inv{\sin}\lrpar{\dfrac{\sqrt{2}}{2}}\)
Solution.
\(\inv{\sin}\lrpar{\dfrac{\sqrt{2}}{2}} = \dfrac{\pi}{4}\)
Example 6.6.4.
Evaluate \(\tan\lrpar{\arcsin \dfrac{1}{4}}\)
Solution.
\(\tan\lrpar{\arcsin \dfrac{1}{4}} = \dfrac{1}{\sqrt{15}}\)
Example 6.6.5.
Let \(f(x) = \cos x\)
  1. Graph \(\cos x\) on the interval \([-2\pi,2\pi]\)
  2. Identify an interval on which \(\cos x\) could possess an inverse
  3. Sketch the graph of the inverse function
Solution.
  1. The graph of \(\cos x\) on the interval \([-2\pi,2\pi]\)
  2. Cosine achieves all its potential values on the interval \([0,\pi]\text{,}\) so use this as our restricted domain.
  3. The graph of \(\arccos x\)
Inverse Cosine (Arccosine).
The function \(y = \cos x\) has the inverse \(x=\inv{\cos}y\) (also written \(\arccos y\)). The domain of \(\inv{\cos}y\) is \([-1,1]\) and the range is \([0,\pi]\)
Example 6.6.6.
Repeat the process we used for sine and cosine to find an inverse function for tangent. Sketch the graph of the inverse function.
Solution.
The graph of tangent is below:
The graph of \(\tan x\) on the interval \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\)
From the graph, we can identify a potential restricted domain as \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\)
The graph of this inverse is given below:
The graph of \(\arctan x\) on \([-10,10]\)
Inverse Tangent (Arctangent).
The function \(y = \tan x\) has the inverse \(x=\inv{\tan}y\) (also written \(\arctan y\)). The domain of \(\inv{\tan}y\) is \((-\infty,\infty)\) and the range is \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\)
Example 6.6.7.
Simplify the expression \(\tan\lrpar{\arccos x}\)
Solution.
\(\tan\lrpar{\arccos x} = \dfrac{\sqrt{1-x^2}}{x}\)
Example 6.6.8.
Using the sketch of arctangent from above, compute \(\ds\lim_{x\to \infty} \arctan x\) and \(\ds\lim_{x\to -\infty} \arctan x\)
Solution.
\begin{align*} \ds\lim_{x\to \infty} \arctan x \amp = \dfrac{\pi}{2}\\ \ds\lim_{x\to -\infty} \arctan x \amp = -\dfrac{\pi}{2} \end{align*}

Subsection 6.6.2 Pre-Class Activities

Example 6.6.9.

Evaluate the following:
  1. \(\displaystyle \inv{\tan}(\sqrt{3})\)
  2. \(\displaystyle \inv{\cos}\lrpar{\dfrac{\sqrt{3}}{2}}\)
  3. \(\displaystyle \tan\lrpar{\arcsin\lrpar{\dfrac{2}{3}}}\)
  4. \(\displaystyle \cos\lrpar{2\inv{\sin}\lrpar{\dfrac{5}{13}}}\)
Solution.
  1. \(\displaystyle \dfrac{\pi}{3}\)
  2. \(\displaystyle \dfrac{\pi}{6}\)
  3. \(\displaystyle \dfrac{2}{\sqrt{5}}\)
  4. \(\displaystyle \dfrac{119}{169}\)

Example 6.6.10.

Simplify the expressions:
  1. \(\displaystyle \tan \lrpar{\inv{\sin}(x)}\)
  2. \(\displaystyle \cos\lrpar{\inv{\sin}(x)}\)
  3. \(\displaystyle \sin (2\arccos x)\)
Solution.
  1. \(\displaystyle \dfrac{x}{\sqrt{1-x^2}}\)
  2. \(\displaystyle \sqrt{1-x^2}\)
  3. \(\displaystyle x\sqrt{1-x^2}\)

Subsection 6.6.3 In Class

Subsubsection 6.6.3.1 Derivatives of Inverse Trig Functions

Function Domain Range
\(\inv{sin}(x)\) \([-1,1]\) \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\)
\(\inv{\cos}(x)\) \([-1,1]\) \([0,\pi]\)
\(\inv{\tan}(x)\) \((-\infty,\infty)\) \(\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\)
\(\inv{\cot}(x)\) \((-\infty\infty)\) \((0,\pi)\)
\(\inv{\sec}(x)\) \((-\infty,1)\cup (1,\infty)\) \(\left[0,\dfrac{\pi}{2}\right)\cup \left[\pi, \dfrac{3\pi}{2}\right)\)
\(\inv{\csc}(x)\) \((-\infty,1)\cup (1,\infty)\) \(\left(0,\dfrac{\pi}{2}\right]\cup \left(\pi, \dfrac{3\pi}{2}\right]\)
Example 6.6.11.
Use implicit differentiation to show that \(\dfrac{d}{dx}\left[\inv{\sin}(x)\right] = \dfrac{1}{\sqrt{1-x^2}}\)
Solution.
Let \(y = \sin x\text{.}\) Then, \(x = \sin y\text{.}\) Taking the implicit derivative, we have
\begin{equation*} 1 = \cos y \dfrac{dy}{dx} \end{equation*}
From the Pythagorean Theorem, we know that \(\cos^2 y = 1-\sin^2y\text{.}\) But \(x = \sin y\text{,}\) so \(\cos^2 y = 1-x^2 \text{.}\) Then, we can say that \(\cos y = \pm \sqrt{1-x^2}\text{.}\)
From the restricted domain for \(x = \sin y\text{,}\) we know that \(x\in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\text{,}\) so that \(\cos y\) is always positive (the angle is in quadrants 1 and 4). This allows us to say that \(\cos y = \sqrt{1-x^2}\text{.}\)
Solving for \(\dfrac{dy}{dx}\text{,}\) we have
\begin{align*} \dfrac{dy}{dx} \amp = \dfrac{1}{\cos y}\\ \amp = \dfrac{1}{\sqrt{1-x^2}} \end{align*}
Example 6.6.12.
Use the same process to find the derivative of \(\inv{\cos}(x)\)
Solution.
Let \(y = \cos x\text{.}\) Then, \(x = \cos y\text{.}\) Taking the implicit derivative, we have
\begin{equation*} 1 = -\sin y \dfrac{dy}{dx} \end{equation*}
From the Pythagorean Theorem, we know that \(\sin^2 y = 1-\cos^2y\text{.}\) But \(x = \cos y\text{,}\) so \(\sin^2 y = 1-x^2 \text{.}\) Then, we can say that \(\sin y = \pm \sqrt{1-x^2}\text{.}\)
From the restricted domain for \(x = \cos y\text{,}\) we know that \(x\in [0,\pi]\text{,}\) so that \(\sin y\) is always positive (the angle is in quadrants 1 and 2). This allows us to say that \(\sin y = \sqrt{1-x^2}\text{.}\)
Solving for \(\dfrac{dy}{dx}\text{,}\) we have
\begin{align*} \dfrac{dy}{dx} \amp = -\dfrac{1}{\sin y}\\ \amp = -\dfrac{1}{\sqrt{1-x^2}} \end{align*}
So \(\dfrac{d}{dx}[\inv{\cos}(x)] = -\dfrac{1}{\sqrt{1-x^2}}\)
Example 6.6.13.
Again, use the same process to find the derivative of \(\inv{\tan}(x)\)
Solution.
Let \(y = \tan x\text{.}\) Then, \(x = \tan y\text{.}\) Taking the implicit derivative, we have
\begin{equation*} 1 = \sec^2 y \dfrac{dy}{dx} \end{equation*}
From the Pythagorean Theorem, we know that \(\sec^2 y = 1+\tan^2y\text{.}\) Since \(x=\tan y\text{,}\) we conclude that \(\sec^2y = 1 + x^2\)
Solving for \(\dfrac{dy}{dx}\text{,}\) we have
\begin{align*} \dfrac{dy}{dx} \amp = \dfrac{1}{\sec^2 y}\\ \amp = \dfrac{1}{1+x^2} \end{align*}
So \(\dfrac{d}{dx}[\inv{\tan}(x)] = \dfrac{1}{1+x^2}\)
The table below collects the derivatives of the six inverse trig functions:
Function Derivative Function Derivative
\(\inv{\sin}(x)\) \(\dfrac{1}{\sqrt{1-x^2}}\) \(\inv{\csc}(x)\) \(-\dfrac{1}{x\sqrt{x^2-1}}\)
\(\inv{\cos}(x)\) \(-\dfrac{1}{\sqrt{1-x^2}}\) \(\inv{\sec}(x)\) \(\dfrac{1}{x\sqrt{x^2-1}}\)
\(\inv{\tan}(x)\) \(\dfrac{1}{1+x^2}\) \(\inv{\cot}(x)\) \(-\dfrac{1}{1+x^2}\)
Example 6.6.14.
Find the domain of the function \(y = \arcsin(x^2-4)\text{.}\) Then, find its derivative, and the domain of the derivative.
Solution.
The domain is \([-\sqrt{5},-\sqrt{3}]\cup [\sqrt{3},\sqrt{5}]\)
The derivative is \(y' = \dfrac{2x}{\sqrt{1-(x^2-4)^2}}\)
The domain of the derivative is \((-\sqrt{5},-\sqrt{3})\cup (\sqrt{3},\sqrt{5})\)
Example 6.6.15.
Find the derivative of \(f(x) = x\arccos(\sqrt{x})\)
Solution.
\(f'(x) = \arccos(\sqrt{x}) - \dfrac{\sqrt{x}}{2\sqrt{1-x}}\)
Example 6.6.16.
Write the derivative of \(f(x) = \dfrac{1}{\inv{\tan}(x)}\)
Solution.
\(f'(x) = -\dfrac{1}{(\inv{\tan}(x))^2(1+x^2)}\)
Example 6.6.17.
Find the derivatives:
  1. \(\displaystyle \lrpar{\inv{\tan}(x)}^2\)
  2. \(\displaystyle \inv{\cot}(t) + \inv{\cot}\lrpar{\dfrac{1}{t}}\)
Solution.
  1. \(\displaystyle \dfrac{2\arctan x}{1+x^2}\)
  2. \(\displaystyle -\dfrac{1}{1+t^2} + \dfrac{t^{-2}}{1+t^{-2}}\)
Example 6.6.18.
If \(g(x) = x\inv{\sin}\lrpar{\dfrac{x}{4}} +\sqrt{16-x^2}\text{,}\) find \(g'(2)\text{.}\)
Solution.
\(g'(2) = \dfrac{\pi}{6}\)
Example 6.6.19.
Find an equation of the tangent line to the curve \(y = 3\arccos\lrpar{\dfrac{x}{2}}\) at the point \((1,\pi)\text{.}\)
Solution.
\(y = -\dfrac{3}{\sqrt{3}}x + \dfrac{3}{\sqrt{3}}+\pi\)
Example 6.6.20.
Find \(y'\) if \(\inv{\tan}(x^2y) = x + xy^2\)
Solution.
\begin{equation*} \dfrac{dy}{dx} = \dfrac{1+y^2-\dfrac{2xy}{1+(x^2y)^2}}{\dfrac{x^2}{1+(x^2y)^2}-2xy} \end{equation*}

Subsubsection 6.6.3.2 Integrals of Inverse Trig Functions

The derivatives for the inverse trig functions give way to corresponding antiderivatives. There are two important ones:
Function Derivative Function Derivative
\(\dfrac{1}{\sqrt{1-x^2}}\) \(\inv{\sin}(x) + C\) \(\dfrac{1}{1+x^2}\) \(\inv{\tan}(x) + C\)
Example 6.6.21.
Compute \(\ds \int_0^{1/4} \dfrac{1}{\sqrt{1-4x^2}}\, dx\)
Solution.
\(\dfrac{\pi}{12}\)
Example 6.6.22.
For any real number \(a\text{,}\) find \(\ds \int \dfrac{1}{x^2 + a^2}\, dx\)
Solution.
\(\ds \int \dfrac{1}{x^2 + a^2}\, dx = \dfrac{1}{a}\inv{\tan}\lrpar{\dfrac{x}{a}} + C\)

Subsection 6.6.4 After Class Activities

Example 6.6.23.

Compute the derivatives of the functions below:
  1. \(\displaystyle f(x) = x\inv{\sin}(x) + \sqrt{1-x^2}\)
  2. \(\displaystyle F(x) = x\inv{\sec}(x^3)\)
  3. \(\displaystyle y = \inv{\tan}\lrpar{x-\sqrt{1+x^2}}\)
Solution.

Example 6.6.24.

Explain why \(\ds \lim_{x\to\infty} \arccos \lrpar{\dfrac{1+x^2}{1+2x^2}} = \dfrac{\pi}{3}\)
Solution.

Example 6.6.25.

Compute the integrals:
  1. \(\displaystyle \ds \int_{1/\sqrt{3}}^{\sqrt{3}} \dfrac{8}{1+x^2}\, dx\)
  2. \(\displaystyle \ds \int_0^{\pi/2} \dfrac{\sin x}{1+\cos^2x}\, dx\)
  3. \(\displaystyle \ds \int \dfrac{e^{2x}}{\sqrt{1-e^{4x}}}\, dx\)
Solution.
  1. \(\displaystyle \ds \int_{1/\sqrt{3}}^{\sqrt{3}} \dfrac{8}{1+x^2}\, dx = \dfrac{4\pi}{3}\)
  2. \(\displaystyle \ds \int_0^{\pi/2} \dfrac{\sin x}{1+\cos^2x}\, dx = \dfrac{\pi}{4}\)
  3. \(\displaystyle \ds \int \dfrac{e^{2x}}{\sqrt{1-e^{4x}}}\, dx = \dfrac{1}{2}\inv{\sin}\lrpar{e^{2x}}\)
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