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Mathematics LibreTexts

2.8: Limits and continuity of Inverse Trigonometric functions

This page is a draft and is under active development. 

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Inverse functions

Recall that a function f is one-to-one (often written as 11) if it assigns distinct values of y to distinct values of x. In other words, if x1x2 then f(x1)f(x2). Equivalently, f is one-to-one if f(x1)=f(x2) implies x1=x2. There is a simple horizontal rule for determining whether a function y=f(x) is one-to-one: f is one-to-one if and only if every horizontal line intersects the graph of y=f(x) in the xy-coordinate plane at most once (see Figure 5.3.3).

alt
Figure 5.3.3 Horizontal rule for one-to-one functions

If a function f is one-to-one on its domain, then f has an inverse function, denoted by f1, such that y=f(x) if and only if f1(y)=x. The domain of f1 is the range of f.

The basic idea is that f1 "undoes'' what f does, and vice versa. In other words,
f1(f(x)) = xfor all x in the domain of f, andf(f1(y)) = yfor all y in the range of f.

Theorem 2.8.1

If f is continuous and one to one, then \(f^{-1}\ is continuous on its domain.

Inverse Trigonometric functions

We know from their graphs that none of the trigonometric functions are one-to-one over their entire domains. However, we can restrict those functions to subsets of their domains where they are one-to-one. For example, y=sinx is one-to-one over the interval [π2,π2], as we see in the graph below:

alt

For π2xπ2 we have 1sinx1, so we can define the inverse sine function y=sin1x (sometimes called the arc sine and denoted by y=arcsin(x) whose domain is the interval [1,1] and whose range is the interval [π2,π2]. In other words:

sin1(siny) = yfor π2yπ2sin(sin1x) = xfor 1x1

Summary of Inverse Trigonometric functions

Lets illustrate the summary of Trigonometric functions and Inverse Trigonometric functions in following table:

Trigonometric function graph of the Trigonometric function

Restricted domain

and

the range

Inverse Trigonometric function

graph of the Inverse Trigonometric function

Properties
f(x)=sin(x) alt

[π2,π2]

and [1,1]

f1(x)=sin1x alt
f(x)=cos(x) alt

[0,π]

and [1,1]

f1(x)=cos1x alt cos1(cosy) = yfor 0yπcos(cos1x) = xfor 1x1

f(x)=tan(x)

alt

(π2,π2)

and R

f1(x)=tan1x

alt tan1(tany) = yfor π2<y<π2tan(tan1x) = xfor all real x

f(x)=cot(x)

alt cot1(coty) = yfor 0<y<πcot(cot1x) = xfor all real x
f(x)=sec(x)

[0,π], with xπ2

and

R

alt

csc1(cscy) = yfor π2yπ2y0csc(csc1x) = xfor |x|1

sec1(secy) = yfor 0yπyπ2sec(sec1x) = xfor |x|1

Below are examples:

Example 2.8.1:

Find sin1(sinπ4).

Solution

Since π2π4π2, we know that sin1(sinπ4)=π4, by Equation 2.8.2.

Example 2.8.2:

Find sin1(sin5π4).

Solution

Since 5π4>π2, we can not use Equation 2.8.2. But we know that sin5π4=12. Thus, sin1(sin5π4)=sin1(12) is, by definition, the angle y such that π2yπ2 and siny=12. That angle is y=π4, since

sin(π4) = sin(π4) = 12 .

Example 2.8.3:

Find cos1(cosπ3).

Solution

Since 0π3π, we know that cos1(cosπ3)=π3, by Equation 2.8.5.

Example 2.8.4:

Find cos1(cos4π3).

Solution

Since 4π3>π, we can not use Equation 2.8.5. But we know that cos4π3=12. Thus, cos1(cos4π3)=cos1(12) is, by definition, the angle y such that 0yπ and cosy=12. That angle is y=2π3 (i.e. 120). Thus, cos1(cos4π3)=2π3.

Example 2.8.5:

Find tan1(tanπ4).

Solution

Since π2π4π2, we know that tan1(tanπ4)=π4, by Equation 2.8.8.

Example 2.8.6:

Find tan1(tanπ).

Solution

Since π>π2, we can not use Equation 2.8.8. But we know that tanπ=0. Thus, tan1(tanπ)=tan10 is, by definition, the angle y such that π2yπ2 and tany=0. That angle is y=0. Thus, tan1(tanπ)=0.

Example2.8.7:

Find the exact value of cos(sin1(14)).

Solution

Let θ=sin1(14). We know that π2θπ2, so since sinθ=14<0, θ must be in QIV. Hence cosθ>0. Thus,

cos2θ = 1  sin2θ = 1  (14)2 = 1516cosθ = 154 .

Note that we took the positive square root above since cosθ>0. Thus, cos(sin1(14))=154.

Example 2.8.8:

Show that tan(sin1x)=x1x2 for 1<x<1.

Solution

When x=0, the Equation holds trivially, since

tan(sin10) = tan0 = 0 = 0102 .

Now suppose that 0<x<1. Let θ=sin1x. Then θ is in QI and sinθ=x. Draw a right triangle with an angle θ such that the opposite leg has length x and the hypotenuse has length 1, as in Figure 5.3.10 (note that this is possible since 0<x<1). Then sinθ=x1=x. By the Pythagorean Theorem, the adjacent leg has length 1x2. Thus, tanθ=x1x2.

If 1<x<0 then θ=sin1x is in QIV. So we can draw the same triangle except that it would be "upside down'' and we would again have tanθ=x1x2, since the tangent and sine have the same sign (negative) in QIV. Thus, tan(sin1x)=x1x2 for 1<x<1.

alt

Example 2.8.9:

Prove the identity tan1x+cot1x = π2.

Solution:

Let θ=cot1x. Using relations, we have

tan(π2θ) = tan(θπ2) = cotθ = cot(cot1x) = x ,

by Equation 2.8.12. So since tan(tan1x)=x for all x, this means that tan(tan1x)=tan(π2θ). Thus, tan(tan1x)=tan(π2cot1x). Now, we know that 0<cot1x<π, so π2<π2cot1x<π2, i.e. π2cot1x is in the restricted subset on which the tangent function is one-to-one. Hence, tan(tan1x)=tan(π2cot1x) implies that tan1x=π2cot1x, which proves the identity.

Continuity of Inverse Trigonometric functions

Example 2.8.1:

Let f(x)=3sec1(x)4tan1(x). Find the values(if any) for which f(x) is continuous.

Exercise 2.8.1

Let f(x)=3sec1(x)8+2tan1(x). Find the values(if any) for which f(x) is continuous.

Answer

Limit of Inverse Trigonometric functions

Theorem 2.8.1

limxtan1(x)=π2.

limxtan1(x)=π2.

limxsec1(x)=limxsec1(x)=π2.

Example 2.8.1:

Find limxsin(2tan1(x)).

Exercise 2.8.1

Find limxsin(2tan1(x)).

Answer

Contributors and Attributions


2.8: Limits and continuity of Inverse Trigonometric functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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