2.8: Limits and continuity of Inverse Trigonometric functions
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Inverse functions
Recall that a function f is one-to-one (often written as 1−1) if it assigns distinct values of y to distinct values of x. In other words, if x1≠x2 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).
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 f−1, such that y=f(x) if and only if f−1(y)=x. The domain of f−1 is the range of f.
The basic idea is that f−1 "undoes'' what f does, and vice versa. In other words,
f−1(f(x)) = xfor all x in the domain of f, andf(f−1(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:
For −π2≤x≤π2 we have −1≤sinx≤1, so we can define the inverse sine function y=sin−1x (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:
sin−1(siny) = yfor −π2≤y≤π2sin(sin−1x) = xfor −1≤x≤1
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) | ![]() |
[−π2,π2] and [−1,1] |
f−1(x)=sin−1x | ![]() |
|
f(x)=cos(x) | ![]() |
[0,π] and [−1,1] |
f−1(x)=cos−1x | ![]() |
cos−1(cosy) = yfor 0≤y≤πcos(cos−1x) = xfor −1≤x≤1 |
f(x)=tan(x) |
![]() |
(−π2,π2) and R |
f−1(x)=tan−1x |
![]() |
tan−1(tany) = yfor −π2<y<π2tan(tan−1x) = xfor all real x |
f(x)=cot(x) |
![]() |
cot−1(coty) = yfor 0<y<πcot(cot−1x) = xfor all real x | |||
f(x)=sec(x) |
[0,π], with x≠π2 and R |
![]() |
csc−1(cscy) = yfor −π2≤y≤π2, y≠0csc(csc−1x) = xfor |x|≥1 sec−1(secy) = yfor 0≤y≤π, y≠π2sec(sec−1x) = xfor |x|≥1 |
Below are examples:
Example 2.8.1:
Find sin−1(sinπ4).
Solution
Since −π2≤π4≤π2, we know that sin−1(sinπ4)=π4, by Equation 2.8.2.
Example 2.8.2:
Find sin−1(sin5π4).
Solution
Since 5π4>π2, we can not use Equation 2.8.2. But we know that sin5π4=−1√2. Thus, sin−1(sin5π4)=sin−1(−1√2) is, by definition, the angle y such that −π2≤y≤π2 and siny=−1√2. That angle is y=−π4, since
sin(−π4) = −sin(π4) = −1√2 .
Example 2.8.3:
Find cos−1(cosπ3).
Solution
Since 0≤π3≤π, we know that cos−1(cosπ3)=π3, by Equation 2.8.5.
Example 2.8.4:
Find cos−1(cos4π3).
Solution
Since 4π3>π, we can not use Equation 2.8.5. But we know that cos4π3=−12. Thus, cos−1(cos4π3)=cos−1(−12) is, by definition, the angle y such that 0≤y≤π and cosy=−12. That angle is y=2π3 (i.e. 120∘). Thus, cos−1(cos4π3)=2π3.
Example 2.8.5:
Find tan−1(tanπ4).
Solution
Since −π2≤π4≤π2, we know that tan−1(tanπ4)=π4, by Equation 2.8.8.
Example 2.8.6:
Find tan−1(tanπ).
Solution
Since π>π2, we can not use Equation 2.8.8. But we know that tanπ=0. Thus, tan−1(tanπ)=tan−10 is, by definition, the angle y such that −π2≤y≤π2 and tany=0. That angle is y=0. Thus, tan−1(tanπ)=0.
Example2.8.7:
Find the exact value of cos(sin−1(−14)).
Solution
Let θ=sin−1(−14). We know that −π2≤θ≤π2, so since sinθ=−14<0, θ must be in QIV. Hence cosθ>0. Thus,
cos2θ = 1 − sin2θ = 1 − (−14)2 = 1516⇒cosθ = √154 .
Note that we took the positive square root above since cosθ>0. Thus, cos(sin−1(−14))=√154.
Example 2.8.8:
Show that tan(sin−1x)=x√1−x2 for −1<x<1.
Solution
When x=0, the Equation holds trivially, since
tan(sin−10) = tan0 = 0 = 0√1−02 .
Now suppose that 0<x<1. Let θ=sin−1x. 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 √1−x2. Thus, tanθ=x√1−x2.
If −1<x<0 then θ=sin−1x is in QIV. So we can draw the same triangle except that it would be "upside down'' and we would again have tanθ=x√1−x2, since the tangent and sine have the same sign (negative) in QIV. Thus, tan(sin−1x)=x√1−x2 for −1<x<1.
Example 2.8.9:
Prove the identity tan−1x+cot−1x = π2.
Solution:
Let θ=cot−1x. Using relations, we have
tan(π2−θ) = −tan(θ−π2) = cotθ = cot(cot−1x) = x ,
by Equation 2.8.12. So since tan(tan−1x)=x for all x, this means that tan(tan−1x)=tan(π2−θ). Thus, tan(tan−1x)=tan(π2−cot−1x). Now, we know that 0<cot−1x<π, so −π2<π2−cot−1x<π2, i.e. π2−cot−1x is in the restricted subset on which the tangent function is one-to-one. Hence, tan(tan−1x)=tan(π2−cot−1x) implies that tan−1x=π2−cot−1x, which proves the identity.
Continuity of Inverse Trigonometric functions
Example 2.8.1:
Let f(x)=3sec−1(x)4−tan−1(x). Find the values(if any) for which f(x) is continuous.
Exercise 2.8.1
Let f(x)=3sec−1(x)8+2tan−1(x). Find the values(if any) for which f(x) is continuous.
- Answer
Limit of Inverse Trigonometric functions
Theorem 2.8.1
limx→∞tan−1(x)=π2.
limx→−∞tan−1(x)=−π2.
limx→∞sec−1(x)=limx→∞sec−1(x)=π2.
Example 2.8.1:
Find limx→∞sin(2tan−1(x)).
Exercise 2.8.1
Find limx→−∞sin(2tan−1(x)).
- Answer
Contributors and Attributions
Michael Corral (Schoolcraft College). The content of this page is distributed under the terms of the GNU Free Documentation License, Version 1.2.
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.
Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)