7.3: Inverse Trigonometric Functions

Inverse functions were studied in Section 2.4. We now take up the topic again and apply it to trigonometric functions. A binary relation on the real numbers is any set of ordered pairs of real numbers. Thus a real function \(f\) of one variable is a binary relation such that for each \(x\), either there is exactly one \(y\) with \((x, y)\) in \(f\) or there is no \(y\) with \((x, y)\) in \(f\). (Other important relations are \(x<y, x \leq y, x \neq y, x=y\).)

Definition

Let \(S\) be a binary relation on the real numbers. The inverse relation of \(S\) is the set \(T\) of all ordered pairs \((y, x)\) such that \((x, y)\) is in \(S\). If \(S\) and \(T\) are both functions they are called inverse functions of each other.

The inverse of a function \(f\) may or may not be a function. For example, the inverse of \(y=x^{2}\) is the relation \(x= \pm \sqrt{y}\), which is not a function (Figure 7.3.1). But the inverse of \(y=x^{2}, x \geq 0\), is the function \(x=\sqrt{y}\) (Figure 7.3.2).

Figure 7.3 .1

Figure 7.3.2

 

Geometrically, the graph of the inverse relation of \(y=f(x)\) can be obtained by flipping the graph of \(y=f(x)\) about the diagonal line \(y=x\) (the dotted line in Figures 7.3.1 and 7.3.2). This flipping interchanges the \(x\) - and \(y\)-axes. This is because \(f(x)=y\) means \((x, y)\) is in \(f\), and \(g(y)=x\) means \((y, x)\) is in \(g\). It follows that:

If \(f\) and \(g\) are inverse functions then the range of \(f\) is the domain of \(g\) and vice versa.

Which functions have inverse functions? We can answer this question with a definition and a simple theorem.

Definition

A real function \(f\) with domain \(X\) is said to be one-to-one if \(f\) never takes the same value twice, that is, for all \(x_{1} \neq x_{2}\) in \(X\) we have \(f\left(x_{1}\right) \neq f\left(x_{2}\right)\).

Theorem 1

\(f\) has an inverse function if and only if \(f\) is one-to-one.

PROOF

The following statements are equivalent.

1. \(f\) is a one-to-one function.
2. For every \(y\), either there is exactly one \(x\) with \(f(x)=y\) or there is no \(x\) with \(f(x)=y\).
3. The equation \(y=f(x)\) determines \(x\) as a function of \(y\).
4. \(f\) has an inverse function.

Corollary

Every function which is increasing on its domain I has an inverse function. So does every function decreasing on its domain \(I\).

PROOF Let \(f\) be increasing on \(I\). For any two points \(x_{1} \neq x_{2}\) in \(I\), the value of \(f\) at the smaller of \(x_{1}, x_{2}\) is less than the value of \(f\) at the greater, so \(f\left(x_{1}\right) \neq\) \(f\left(x_{2}\right)\).

For example, the function \(y=x^{2}\) is not one-to-one because \((-1)^{2}=1^{2}\), whence it has no inverse function. The function \(y=x^{2}, x \geq 0\), is increasing on its domain \([0, \infty)\) and thus has an inverse.

Now let us examine the trigonometric functions. The function \(y=\sin x\) is not one-to-one. For example, \(\sin 0=0, \sin \pi=0, \sin 2 \pi=0\), etc. We can see in Figure 7.3.3 that the inverse relation of \(y=\sin x\) is not a function.

\(y=\sin x\)

not a function

Figure 7.3.3

However, the function \(y=\sin x\) is increasing on the interval \([-\pi / 2, \pi / 2]\), because its derivative \(\cos x\) is \(\geq 0\). So the sine function restricted to the interval \([-\pi / 2, \pi / 2]\),

\[y=\sin x, \quad-\pi / 2 \leq x \leq \pi / 2 \nonumber \]

has an inverse function shown in Figure 7.3.4. This inverse is called the arcsine

\(y=\sin x,-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\)

Figure 7.3.4
function. It is written \(x=\arcsin y\). Verbally, arcsin \(y\) is the angle \(x\) between \(-\pi / 2\) and \(\pi / 2\) whose sine is \(y\).

The other trigonometric functions also are not one-to-one and thus do not have inverse functions. However, in each case we obtain a one-to-one function by restricting the domain to a suitable interval, either \([-\pi / 2, \pi / 2]\) or \([0, \pi]\). The resulting inverse functions are called the arccosine, arctangent, etc.

Definition

The inverse trigonometric functions are defined as follows.

\(x=\arcsin y\) is the inverse of \(y=\sin x, \quad-\pi / 2 \leq x \leq \pi / 2\)

\(x=\arccos y\) is the inverse of \(y=\cos x, \quad 0 \leq x \leq \pi\)

\(x=\arctan y\) is the inverse of \(y=\tan x, \quad-\pi / 2 \leq x \leq \pi / 2\)

\(x=\operatorname{arccot} y\) is the inverse of \(y=\cot x, \quad 0 \leq x \leq \pi\)

\(x=\operatorname{arcsec} y\) is the inverse of \(y=\sec x, \quad 0 \leq x \leq \pi\)

\(x=\operatorname{arccsc} y\) is the inverse of \(y=\csc x, \quad-\pi / 2 \leq x \leq \pi / 2\)

 

The graphs of these functions are shown in Figure 7.3.5. The domains of the inverse trigonometric functions can be read off from the graphs, and are shown in the table below.

Table 7.3.1

Function	Domain
\(\arcsin y\)	\(-1 \leq y \leq 1\)
\(\arccos y\)	\(-1 \leq y \leq 1\)
\(\arctan y\)	whole real line
\(\operatorname{arccot} y\)	whole real line
\(\operatorname{arcsec} y\)	\(y \leq-1, \quad y \geq 1\)
\(\operatorname{arccsc} y\)	\(y \leq-1, \quad y \geq 1\)

We can prove the inverse trigonometric functions have these domains (i.e., the figures are correct) using the Intermediate Value Theorem. As an illustration we prove that arcsin \(y\) has domain \([-1,1]\).

\(\arcsin y\) is undefined outside \([-1,1]\) because \(-1 \leq \sin x \leq 1\) for all \(x\). Suppose \(y_{0}\) is in \([-1,1]\). Then

\[\sin (-\pi / 2)=-1 \leq y_{0} \leq 1=\sin (\pi / 2) \nonumber \]

\(\sin x\) is continuous, so by the Intermediate Value Theorem there exists \(x_{0}\) between \(-\pi / 2\) and \(\pi / 2\) such that \(\sin x_{0}=y_{0}\). Thus

\[\arcsin y_{0}=x_{0} \nonumber \]

and \(y_{0}\) is in the domain of arcsin \(y\).

Example 1

Find \(\arccos (\sqrt{2} / 2)\). From Table 7.1.1, \(\cos (\pi / 4)=\sqrt{2} / 2\). Since \(0 \leq \pi / 4 \leq \pi\),

Solution

\[\arccos (\sqrt{2} / 2)=\pi / 4 \nonumber \]

 

Figure 7.3.5

 

Example 2

Find \(\arcsin (-1)\). From Table 7.1.1, \(\sin (3 \pi / 2)=-1\). But \(3 \pi / 2\) is not in the interval \([-\pi / 2, \pi / 2]\). Using \(\sin (\theta+2 n \pi)=\sin \theta\), we have

Solution

\[\begin{aligned} \sin (-\pi / 2) & =\sin (3 \pi / 2)=-1 \\ \arcsin (-1) & =-\pi / 2 \end{aligned} \nonumber \]

Example 3

Find \(\arctan (-\sqrt{3})\). We must find a \(\theta\) in the interval \([-\pi / 2, \pi / 2]\) such that \(\tan \theta=-\sqrt{3}\). From Table 7.1.1, \(\sin (\pi / 3)=\sqrt{3} / 2, \cos (\pi / 3)=1 / 2\). Then \(\sin (-\pi / 3)=-\sqrt{3} / 2, \cos (-\pi / 3)=1 / 2\). So

Solution

\[\begin{aligned} \tan (-\pi / 3) & =\frac{-\sqrt{3} / 2}{1 / 2}=-\sqrt{3} \\ \arctan (-\sqrt{3}) & =-\pi / 3 \end{aligned} \nonumber \]

Example 4

. Find \(\cos (\arctan y)\). Let \(\theta=\arctan y\). Thus \(\tan \theta=y\). Using

Solution

\[\begin{aligned} \sin ^{2} \theta+\cos ^{2} \theta & =1 \\ \frac{\sin \theta}{\cos \theta} & =y \end{aligned} \nonumber \]

we solve for \(\cos \theta\).

\[\begin{array}{ll} \sin \theta=y \cos \theta, & (y \cos \theta)^{2}+\cos ^{2} \theta=1 \\ \cos ^{2} \theta\left(y^{2}+1\right)=1, & \cos ^{2} \theta \doteq \frac{1}{y^{2}+1} \\ \cos \theta= \pm \frac{1}{\sqrt{y^{2}+1}} \end{array} \nonumber \]

By definition of arctan \(y\), we know that \(-\pi / 2 \leq \theta \leq \pi / 2\). In this interval, \(\cos \theta \geq 0\). Therefore

\[\cos \theta=\frac{1}{\sqrt{y^{2}+1}} \nonumber \]

Example 5

Show that \(\arcsin y+\arccos y=\pi / 2\) (Figure 7.3.6). Let \(\theta=\arcsin y\). We have \(y=\sin \theta=\cos (\pi / 2-\theta)\). Also, when \(-\pi / 2 \leq \theta \leq \pi / 2\), we have

Solution

\[\pi / 2 \geq-\theta \geq-\pi / 2, \quad \pi \geq \pi / 2-\theta \geq 0 \nonumber \]

Thus

\[\pi / 2-\theta=\arccos y \nonumber \]

\[\arcsin y+\arccos y=\theta+(\pi / 2-\theta)=\pi / 2 \nonumber \]

Figure 7.3.6

We shall next study the derivatives of the inverse trigonometric functions. Here is a general theorem which tells us when the derivative of the inverse function exists and gives a rule for computing its value.

INVERSE FUNCTION THEOREM

Suppose a real function \(f\) is differentiable on an open interval \(I\) and \(f\) has an inverse function \(g\). Let \(x\) be a point in \(I\) where \(f^{\prime}(x) \neq 0\) and let \(y=f(x)\). Then

1. \(g^{\prime}(y)\) exists,
2. \(g^{\prime}(y)=\frac{1}{f^{\prime}(x)}\).

We omit the proof that \(g^{\prime}(y)\) exists. Intuitively, the curve \(y=f(x)\) has a nonhorizontal tangent line, so the curve \(x=g(y)\) should have a nonvertical tangent line and thus \(g^{\prime}(y)\) should exist.

The Inverse Function Rule from Chapter 2 says that (ii) is true if we assume (i). The proof of (ii) from (i) is an application of the Chain Rule:

\[g(f(x))=x, \quad g^{\prime}(f(x)) f^{\prime}(x)=1, \quad g^{\prime}(y) f^{\prime}(x)=1, \quad g^{\prime}(y)=\frac{1}{f^{\prime}(x)} \nonumber \]

The Inverse Function Theorem shows that all the inverse trigonometric functions have derivatives. We now evaluate these derivatives.

Theorem 2

(i) \(d(\arcsin x)=\frac{d x}{\sqrt{1-x^{2}}} \quad\) (where \(-1<x<1\).

\[d(\arccos x)=-\frac{d x}{\sqrt{1-x^{2}}} \quad(\text { where }-1<x<1) \nonumber \]

(ii) \(d(\arctan x)=\frac{d x}{1+x^{2}}\).

\[d(\operatorname{arccot} x)=-\frac{d x}{1+x^{2}} \nonumber \]

(iii) \(d(\operatorname{arcsec} x)=\frac{d x}{|x| \sqrt{x^{2}-1}} \quad\) (where \(|x|>1\) ).

\[d(\operatorname{arccsc} x)=-\frac{d x}{|x| \sqrt{x^{2}-1}} \quad(\text { where }|x|>1) \nonumber \]

PROOF

We prove the first part of (i) and (iii). Since the derivatives exist we may use implicit differentiation.

(i) Let \(y=\arcsin x\). Then

\[\begin{aligned} x & =\sin y, \quad-\pi / 2 \leq y \leq \pi / 2 \\ d x & =\cos y d y \end{aligned} \nonumber \]

From \(\sin ^{2} y+\cos ^{2} y=1\) we get

\[\cos y= \pm \sqrt{1-\sin ^{2} y}= \pm \sqrt{1-x^{2}} \nonumber \]

Since \(-\pi / 2 \leq y \leq \pi / 2, \cos y \geq 0\). Then

\[\cos y=\sqrt{1-x^{2}} \nonumber \]

Substituting,

\[d x=\sqrt{1-x^{2}} d y, \quad d y=\frac{d x}{\sqrt{1-x^{2}}} \nonumber \]

(iii) Let \(y=\operatorname{arcsec} x\).

Then

\[\begin{aligned} x & =\sec y ; \quad 0 \leq y \leq \pi, \\ d x & =\sec y \tan y d y . \end{aligned} \nonumber \]

From \(\tan ^{2} y+1=\sec ^{2} y\) we get \(\tan y= \pm \sqrt{\sec ^{2} y-1}= \pm \sqrt{x^{2}-1}\).

Since \(0 \leq y \leq \pi, \tan y\) and \(\sec y=\frac{1}{\cos y}\) have the same sign.

Therefore

\(\sec y \tan y^{\prime} \geq 0\)

and

\[\begin{aligned} & d x=|\sec y||\tan y| d y=|x| \sqrt{x^{2}-1} d y \\ & d y=\frac{d x}{|x| \sqrt{x^{2}-1}} \end{aligned} \nonumber \]

When we turn these formulas for derivatives around we get some surprising new integration formulas.

Theorem 3

1. \(\int \frac{1}{\sqrt{1-x^{2}}} d x=\arcsin x+C=-\arccos x+C\).(Provided that \(|x|<1\) ).
2. \(\int \frac{d x}{1+x^{2}}=\arctan x+C=-\operatorname{arccot} x+C\).
3. \(\int \frac{d x}{|x| \sqrt{x^{2}-1}}=\operatorname{arcsec} x+C=-\operatorname{arccsc} x+C\). (Provided that \(|x|>1\) ).

From part (i), arcsin \(x\) and -arccos \(x\) must differ only by a constant. We already knew this from Example 5,

\[\arcsin x=-\arccos x+\pi / 2 \nonumber \]

Before now we were not able to find the area of the regions under the curves

\[y=\frac{1}{\sqrt{1-x^{2}}}, \quad y=\frac{1}{1+x^{2}}, \quad y=\frac{1}{x \sqrt{x^{2}-1}} \nonumber \]

It is a remarkable and quite unexpected fact that these areas are given by inverse trigonometric functions.

Example 6

(a) Find the area of the region under the curve

\[y=\frac{1}{1+x^{2}} \nonumber \]

for \(-1 \leq x \leq 1\).

(b) Find the area of the region under the same curve for \(-x<x<x\). The regions are shown in Figure 7.3.7.

Solution

(a) \(\left.A=\int_{-1}^{1} \frac{1}{1+x^{2}} d x=\arctan x\right]_{-1}^{1}=\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)=\frac{\pi}{2}\).

Figure 7.3.7

(b) \(A=\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} d x=\int_{-\infty}^{0} \frac{1}{1+x^{2}} d x+\int_{0}^{\infty} \frac{1}{1+x^{2}} d x\)

\[=\lim _{a \rightarrow-\infty} \int_{a}^{0} \frac{1}{1+x^{2}} d x+\lim _{b \rightarrow \infty} \int_{0}^{b} \frac{1}{1+x^{2}} d x \nonumber \]

\[\begin{aligned} & =\lim _{a \rightarrow-\infty}(\arctan 0-\arctan a)+\lim _{b \rightarrow \infty}(\arctan b-\arctan 0) \\ & =-\lim _{a \rightarrow-\infty} \arctan a+\lim _{b \rightarrow \infty} \arctan b \end{aligned} \nonumber \]

From the graph of \(\arctan x\) we see that the first limit is \(-\pi / 2\) and the second limit is \(\pi / 2\), so

\[A=-\left(-\frac{\pi}{2}\right)+\frac{\pi}{2}=\pi \nonumber \]

Thus the region under \(y=1 /\left(1+x^{2}\right)\) has exactly the same area as the unit circle, and half of this area is between \(x=-1\) and \(x=1\).

Example 7

Find \(\int_{-2}^{-\sqrt{2}} \frac{1}{x \sqrt{x^{2}-1}} d x\).

Solution

\[\begin{aligned} \int_{-2}^{-\sqrt{2}} \frac{1}{x \sqrt{x^{2}-1}} d x & =\int_{-2}^{-\sqrt{2}}-\frac{1}{|x| \sqrt{x^{2}-1}} d x \\ & =-\operatorname{arcsec} x]_{-2}^{-\sqrt{2}}=-(\operatorname{arcsec}(-\sqrt{2})-\operatorname{arcsec}(-2)) \\ & =-\left(\frac{3 \pi}{4}-\frac{2 \pi}{3}\right)=-\frac{\pi}{12} \end{aligned} \nonumber \]

PROBLEMS FOR SECTION 7.3

In Problems 1-9, evaluate the given expression.

\(\mathbf{1}\)	\(\arcsin (\sqrt{3} / 2)\)	\(\mathbf{2}\)	\(\arcsin (-1 / 2)\)
\(\mathbf{3}\)	\(\arctan (-1)\)	\(\mathbf{4}\)	\(\sec (\arctan (-1))\)
\(\mathbf{5}\)	\(\operatorname{arcsec} 2\)	6	\(\arcsin (\cos \pi)\)
\(\mathbf{7}\)	\(\sin (\arccos x)\)	8	\(\cot (\operatorname{arcsec} x)\)
9	\(\arcsin (\cos x), \quad 0 \leq x \leq \pi\)

10 Prove the identity \(\arctan (-x)=-\arctan x\).

11 Prove arctan \((1 / x)=\operatorname{arccot} x\), for \(0<x\).

12 Prove \(\arccos (-x)=\pi-\arccos x\)

13 Prove \(\arctan x+\operatorname{arccot} x=\pi / 2\).

Find the derivatives in Problems 14-25.

\(14 y=\arcsin (x / 2)\) 15
\(16 \quad y=(\arcsin x)^{2}\)
\(18 y=\arctan \sqrt{x}\)
\(20 \quad y=x \operatorname{arcsec} x\)
\(22 \quad y^{\prime}=\arccos x+\left(x / \sqrt{1-x^{2}}\right)\)
\(24 \quad u=\operatorname{arcsec} t+\sqrt{t^{2}-1}\)

\[\begin{aligned} & y=\operatorname{arcsec}(5 x-2) \\ & y=\arcsin \left(x^{2}\right) \tag{17}\\ & s=t \arcsin t \\ & y=\arcsin x+\sqrt{1-x^{2}} \\ & y=x \arcsin x+\sqrt{1-x^{2}} \\ & y=\arctan (1 / \sqrt{x}) \tag{25}\end{aligned} \]

26 Evaluate \(\lim _{x \rightarrow \infty} \operatorname{arccsc} x\).
27 Evaluate \(\lim _{x \rightarrow-\infty} \arctan x\).
28 Evaluate \(\lim _{x \rightarrow 0} \frac{\arcsin x}{x}\).
29 Evaluate \(\lim _{x \rightarrow \infty} \frac{\operatorname{arccot} x}{\operatorname{arccsc} x}\).

In Problems 30-47 evaluate the integrals.
30 \(\int \frac{d x}{1+4 x^{2}}\)
31 \(\int \frac{d x}{9+x^{2}}\)
\(32 \int \frac{d x}{\sqrt{4-x^{2}}}\)
\(33 \int \frac{d x}{\sqrt{x-x^{2}}}\)
\(34 \int \frac{\cos x}{1+\sin ^{2} x} d x\)
35
\(\int \frac{d x}{x \sqrt{4 x^{2}-1}}, x>1\)
\(36 \quad \int \frac{x d x}{x^{4}+1}\)
\(\int \frac{x d x}{\sqrt{1-x^{4}}}\)
\(38 \quad \int \frac{d x}{(1+x) \sqrt{x}}\)
\(\int \frac{d x}{x \sqrt{x-1}}\)
40

\[\int \frac{\arctan x}{1+x^{2}} d x \tag{39} \]

\[\int \frac{\arcsin x}{\sqrt{1-x^{2}}} d x \tag{41} \]

\(42 \quad \int_{-\sqrt{3}}^{\sqrt{3}} \frac{d x}{1+x^{2}}\)

\[\int_{1}^{2} \frac{1}{x \sqrt{x^{2}-1}} d x \tag{43} \]

44

\[\int_{-\sqrt{2}}^{-1} \frac{1}{x \sqrt{x^{2}-1}} d x \tag{45} \]

\[\int_{0}^{1 / 2} \frac{1}{\sqrt{1-x^{2}}} d x \nonumber \]

46

\(\int_{0}^{\infty} \frac{d x}{25 x^{2}+1}\)

47

\[\int_{-\infty}^{\infty} \frac{d x}{a^{2}+x^{2}} \nonumber \]

48 Find the area of the region bounded by the \(x\)-axis and the curve \(y=1 / \sqrt{1-x^{2}}\), \(-1<x<1\).

49 Find the area of the region under the curve \(y=1 /\left(x \sqrt{x^{2}-1}\right), 1 \leq x<\infty\).

50 Find the area of the region bounded below by the line \(y=\frac{1}{2}\) and above by the curve \(y=1 /\left(x^{2}+1\right)\).