Inverse trigonometric functions

Section 2.21 Inverse trigonometric functions

Objectives

Define the inverse trigonometric functions.
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Understand the graphical properties of inverse trigonometric functions.
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Derive derivative formulas for the inverse trigonometric functions using our inverse function derivative rule.
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The trigonometric functions are fail to be one-to-one in a spectacular fashion: letting \(f\) be any one of our six trigonometric functions, using the periodicity of these functions, we have

\begin{align*} \dots=f(x-2\pi)=f(x) \amp =f(x+2\pi)=f(x+4\pi)=\dots\text{.} \end{align*}

However, by restricting each of these functions to a certain interval where it is monotonic, and hence one-to-one, we obtain an invertible function.

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Definition 2.21.1. Inverse trigonometric functions.

The following are examples of what are called inverse trigonometric functions.

On the restricted domain \([-\pi/2, \pi/2]\) the function \(f(x)=\sin x\) is one-to-one, with range \([-1,1]\text{.}\) The inverse function of \(f\) restricted to this domain is called the arcsine function, denoted \(f^{-1}(x)=\arcsin x\text{.}\)
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On the restricted domain \([0, \pi]\) the function \(g(x)=\cos x\) is one-to-one, with range \([-1,1]\text{.}\) The inverse function of \(g\) restricted to this domain is called the arccosine function, denoted \(g^{-1}(x)=\arccos x\text{.}\)
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On the restricted domain \((-\pi/2, \pi/2)\) the function \(h(x)=\tan x\) is one-to-one, with range \((-\infty, \infty)\text{.}\) The inverse function of \(h\) restricted to this domain is called the arctangent function, denoted \(h^{-1}(x)=\arctan x\text{.}\)
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Remark 2.21.2.

Occasionally an alternative notation is used to denote inverse trigonometric functions: namely,

\begin{align*} \arcsin x \amp = \sin^{-1}x \amp \arccos x\amp =\cos^{-1} x \amp \arctan x\amp=\tan^{-1} x\text{.} \end{align*}

We will avoid this notation as it misleadingly suggests that the inverse trigonometric functions are reciprocals of the corresponding trigonometric functions. They are not!

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The usual connections between an invertible function \(f\) and its inverse \(f^{-1}\) allow us to deduce properties of the inverse trigonometric functions from their paired trigonometric function.

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Theorem 2.21.3. Properties of inverse trigonometric functions.

The function \(\arcsin\) is an increasing function with domain \([-1,1]\) and range \([0,\pi]\text{.}\) It satisfies the following properties:

\begin{align} \arcsin(x)=\theta \amp \iff \sin\theta=x \text{ and } -\pi/2\leq \theta\leq \pi/2 \tag{2.117}\\ \arcsin(\sin \theta)\amp =\theta \text{ for all } -\pi/2\leq \theta\leq \pi/2\tag{2.118}\\ \sin(\arcsin x)\amp =x \text{ for all } -1\leq x\leq 1\text{.}\tag{2.119} \end{align}

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The function \(\arccos\) is a decreasing function with domain \([-1,1]\) and range \([0,\pi]\text{.}\) It satisfies the following properties:

\begin{align} \arccos(x)=\theta \amp \iff \cos\theta=x \text{ and } 0\leq \theta\leq \pi\tag{2.120}\\ \arccos(\cos \theta)\amp =\theta \text{ for all } 0\leq \theta\leq \pi\tag{2.121}\\ \cos(\arccos x)\amp =x \text{ for all } -1\leq x\leq 1\text{.}\tag{2.122} \end{align}

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The function \(\arctan\) is an increasing function with domain \((-\infty, \infty)\) and range \((-\pi/2, \pi/2)\text{.}\) It satisfies the following properties:

\begin{align} \arctan(x)=\theta \amp \iff \tan\theta=x \text{ and } -\pi/2\lt \theta\lt \pi/2 \tag{2.123}\\ \arctan(\tan \theta)\amp =\theta \text{ for all } -\pi/2\lt \theta\lt \pi/2\tag{2.124}\\ \tan(\arctan x)\amp =x \text{ for all } x\tag{2.125}\\ \lim_{x\to\infty}\arctan x\amp =\pi/2,\hspace{5pt} \lim_{x\to-\infty}\arctan x=-\pi/2\tag{2.126} \end{align}

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Example 2.21.4. Computing with inverse trig functions.

Compute the following values of trigonometric functions by hand.

\(\displaystyle \displaystyle\arcsin(-1)\)
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\(\displaystyle \displaystyle\arccos(-\sqrt{2}/2)\)
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\(\displaystyle \displaystyle\arctan(-1/\sqrt{3})\)
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\(\displaystyle\arcsin\left(\sin\left(\frac{10\pi}{11}\right)\right)\) Hint. The answer is not \(10\pi/11\text{.}\)
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Solution.

Common mistake: \(\arcsin(-1)\ne 3\pi/2\text{.}\) Why? By definition, the outputs of \(\arcsin\) lie in the interval \([-\pi/2, \pi/2]\text{.}\) Instead, using (2.117), we have \(\arcsin(-1)=\theta\) if and only if \(\sin \theta=-1\) and \(-\pi/2\leq \theta\leq \pi/2\text{.}\) The unique \(\theta\) satisfying these two conditions is \(\theta=-\pi/2\text{.}\) We conclude that \(\arcsin(-1)=-\pi/2\text{.}\)
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Using (2.117), we have \(\arccos(-\sqrt{2}/2)=\theta\) if and only if \(\cos \theta=-\sqrt{2}/2\) and \(0/\leq \theta\leq \pi\text{.}\) The unique \(\theta\) satisfying these two conditions is \(\theta=3\pi/4\text{.}\) We conclude that \(\arccos(-\sqrt{2}/2)=3\pi/4\text{.}\)
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Using (2.123), we have \(\arctan(-1/\sqrt{3})=\theta\) if and only if \(\tan \theta=-1/\sqrt{3}\) and \(-\pi/2 < \theta < \pi/2\text{.}\) Since \(\theta=-\pi/6\) is the unique angle satifying these two conditions, we conclude that that \(\arctan(-1/\sqrt{3})=-\pi/6\text{.}\)
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Recall that \(\sin(\theta)=\sin (\pi-\theta)\) for any \(\theta\in \R\text{.}\) (You can convince yourself of this either by drawing a unit circle picture, or by using the difference identity for \(\sin\text{.}\)) It follows that

\begin{align*} \arcsin(\sin(10\pi/11)) \amp =\arcsin(\sin(\pi-10\pi/11))\\ \amp = \arcsin(\sin (\pi /11))\\ \amp = \frac{\pi}{11} \amp (\knowl{./knowl/xref/eq_arcsin_sin.html}{\text{(2.118)}}, -\frac{\pi}{2}\leq \frac{\pi}{11}\leq\frac{\pi}{2})\text{.} \end{align*}

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The inverse trigonometric functions are often used to provide a single solution to an elementary trigonometric equation that we cannot solve “by hand”: i.e., equations whose solutions \(\theta\) do not correspond to one of the “familiar” angles of the unit circle. For example, \(\theta_0=\arcsin(1/3)\) is a particular solution to the equation \(\sin \theta=\frac{1}{3}\text{.}\) But what if we are asked to find all solutions to such an equation? In this case we make use of Procedure 2.19.5.

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Example 2.21.5. Solving trig equations.

Find all solutions to the following trigonometric equations lying within the interval \([0,2\pi]\text{.}\) You may express your answer in terms of values of inverse trigonometric functions.

\(\displaystyle \displaystyle 3\sin 2\theta +4=6\)
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\(\displaystyle \displaystyle \tan (\theta+\pi)=-10\)
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Solution.

After some algebra we rewrite the equation as

\begin{equation*} \sin 2\theta=\frac{2}{3} \end{equation*}

or

\begin{equation} \sin u=\frac{2}{3}\text{,}\tag{2.127} \end{equation}

where \(u=2\theta\text{.}\) Since \(u_0=\arcsin(2/3)\) is a solution to (2.127), according to Procedure 2.19.5 the general solution is

\begin{equation*} u=\arcsin(2/3)+2\pi n \text{ or } u=(\pi-\arcsin(2/3))+2\pi n\text{.} \end{equation*}

Since \(u=2\theta\text{,}\) we conclude that the general solution to the original equation is

\begin{equation*} \theta=\frac{1}{2}\arcsin(2/3)+\pi n \text{ or } \theta=\frac{1}{2}(\pi-\arcsin(2/3))+\pi n\text{.} \end{equation*}

Which of these infinitely-many solutions lies in the prescribed interval \([0,2\pi]\text{?}\) Since \(\arcsin(2/3)\in [0,\pi/2]\text{,}\) we have \(\frac{1}{2}\arcsin(2/3)\in [0,\pi/4]\) and \(\frac{1}{2}(\pi-\arcsin(2/3)\in [\pi/4, \pi/2]\text{.}\) It follows that the solutions lying in \([0,2\pi]\) are

\begin{equation*} \theta=\frac{1}{2}\arcsin(2/3), \frac{1}{2}\arcsin(2/3)+\pi, \frac{1}{2}(\pi-\arcsin(2/3)), \frac{1}{2}(\pi-\arcsin(2/3))+\pi\text{.} \end{equation*}

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Since \(\theta_0=\arctan(-10)-\pi\) is one solution to the given equation, the general solution is given by

\begin{equation*} \theta=\arctan(-10)-\pi +\pi n= \arctan(-10)+(n-1)\pi \text{,} \end{equation*}

where \(n\) is any integer. Which of these lie in the interval \([0,2\pi]\text{?}\) Since \(\arctan(-10)\in (-\pi/2, 0]\text{,}\) we have \(\arctan(-10)+\pi\in (\pi/2, \pi]\) and \(\arctan(-10)+2\pi \in (3\pi/2, 2\pi]\text{.}\) Thus the solutions \(\theta\) lying in \([0,2\pi]\) are \(\theta=\arctan(-10)+\pi\) and \(\theta=\arctan(-10)+2\pi\text{.}\)

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Lastly, we can deduce derivative formulas for \(\arcsin\text{,}\) \(\arccos\text{,}\) and \(\arctan\) using the inverse function derivative rule:

\begin{align*} \frac{d}{dx}(f^{-1}(x)) \amp = \frac{1}{f'(f^{-1}(x))} \text{.} \end{align*}

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Theorem 2.21.6. Derivatives of inverse trig functions.

\begin{align} \frac{d}{dx} \arcsin x\amp =\frac{1}{\sqrt{1-x^2}} \text{ for all } x\in (-1,1)\tag{2.128}\\ \frac{d}{dx} \arccos x\amp =-\frac{1}{\sqrt{1-x^2}} \text{ for all } x\in (-1,1)\tag{2.129}\\ \frac{d}{dx} \arctan x\amp =\frac{1}{1+x^2}\amp \text{ for all } x\in \R \text{.}\tag{2.130} \end{align}

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We prove (2.128) and (2.123); the proof of (2.129) proceeds along similar lines.

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Proof of (2.128).

Let \(f(x)=\sin x\) restricted to the domain \(D=[-\pi/2, \pi/2]\text{,}\) and let \(f^{-1}(x)=\arcsin x\text{.}\) Using Theorem 2.13.1, we have

\begin{align*} \frac{d}{dx} \arcsin x \amp = \frac{1}{f'(\arcsin x)}\\ \amp = \frac{1}{\cos(\arcsin x)}\\ \amp = \frac{1}{\sqrt{1-x^2}}\text{.} \end{align*}

The last step here is perhaps the most interesting, and follows from the fact that \(\cos(\arcsin x)=\sqrt{1-x^2}\) for any \(x\in [-1,1]\text{.}\) Why is this true? Set \(\arcsin x=\theta\text{,}\) an angle lying in \([-\pi/2,\pi/2]\text{.}\) Using the fact that \(\sin \theta=x\text{,}\) we can draw a unit circle picture like the one below. Since the vertical leg of the right triangle there has length \(\abs{x}\text{,}\) the horizontal leg has length \(\sqrt{1-\abs{x}^2}=\sqrt{1-x^2}\text{.}\) Since this length is equal to the \(x\)-coordinate of the point \(P\) below, we have \(\cos \theta=\sqrt{1-x^2}\) and thus \(\cos(\arcsin x)=\sqrt{1-x^2}\text{,}\) as claimed.

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Unit circle picture for arcsin derivative proof

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Proof of (2.130).

Let \(f(x)=\tan x\) restricted to the domain \(D=(-\pi/2, \pi/2)\text{,}\) and let \(f^{-1}(x)=\arctan x\text{.}\) Using Theorem 2.13.1, we have

\begin{align*} \frac{d}{dx}\arctan x \amp = \frac{1}{f'(\arctan x)}\\ \amp = \frac{1}{\sec^2 (\arctan x)}\\ \amp = \frac{1}{(\tan(\arctan x))^2+1} \amp (\tan^2 x+1=\sec^2x)\\ \amp = \frac{1}{x^2+1} \amp (\tan(\arctan x)=x)\text{.} \end{align*}

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Example 2.21.7. Derivatives of inverse trig functions.

Find the equation of the tangent line to \(f(x)=\arccos x\) at \(x=1/2\text{.}\)

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Solution.

The tangent line passes through the point

\begin{equation*} P=(1/2, \arccos(1/2))=(1/2, \pi/3) \end{equation*}

and has slope

\begin{equation*} m=f'(1/2)=-\frac{1}{\sqrt{1-(1/2)^2}}=-\frac{2}{\sqrt{3}}\text{.} \end{equation*}

Using point-slope form, we see that the line has equation

\begin{equation*} y-\frac{\pi}{3}=\frac{-2}{\sqrt{3}}\left(x-\frac{1}{2}\right)\text{.} \end{equation*}

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Example 2.21.8. Limit computation.

Compute \(\displaystyle\lim_{x\rightarrow 1^{-}}\frac{\arccos(x^2)}{\sqrt{1-x}}\text{.}\)

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Solution.

Since \(\arccos(1)=0\text{,}\) we have

\begin{align*} \lim_{x\rightarrow 1^{-}}\frac{\arccos(x^2)}{\sqrt{1-x}}\amp = \lim_{x\rightarrow 1^{-}}\frac{(\arccos(x^2))'}{(\sqrt{1-x})'} \amp (\text{L'Hop }, 0/0) \\ \amp = \lim_{x\rightarrow 1^{-}}\frac{-2x/\sqrt{1-x^4}}{-1/2\sqrt{1-x}}\\ \amp =\lim_{x\rightarrow 1^{-}}4x\sqrt{\frac{1-x}{1-x^4}}\\ \amp =\lim_{x\rightarrow 1^{-}}4x\sqrt{\frac{1-x}{(1-x^2)(1+x^2)}}\\ \amp =\lim_{x\rightarrow 1^{-}}4x\sqrt{\frac{1-x}{(1-x)(1+x)(1+x^2)}}\\ \amp =\lim_{x\rightarrow 1^{-}}4x\sqrt{\frac{1}{(1+x)(1+x^2)}}\\ \amp = 4\cdot \frac{1}{2}\\ \amp = 2\text{.} \end{align*}

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