/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Explain why the domain of the si... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 9

Explain why the domain of the sine function must be restricted in order to define its inverse function.

Short Answer

Expert verified
Answer: The domain of the sine function must be restricted to ensure that it is one-to-one and onto over its range, which are necessary properties for a function to have an inverse. A common restricted domain is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), which satisfies these properties, allowing for the definition of an inverse function.

Step by step solution

01

Understand the inverse function properties

A function has an inverse function if and only if it is both one-to-one (injective) and onto (surjective). In a one-to-one function, each element of the domain maps to a unique element of the codomain. This means that there cannot be two different inputs that yield the same output. In other words, when a function is one-to-one, if \(f(x_1)=f(x_2)\) then \(x_1=x_2\). A function is onto if every element in the codomain has an element from the domain that maps to it. This means that the range of the function is equal to its codomain.
02

Analyze the sine function

The sine function, usually denoted as \(\sin(x)\), is periodic with a period of \(2\pi\). This means that for any value \(x\), we have \(\sin(x) = \sin(x+2n\pi)\) where \(n\) is an integer. It also has a well-defined range: all values of the sine function lie within the interval \([-1, 1]\).
03

Determine if the sine function is one-to-one over its domain

The sine function is not one-to-one over its entire domain, since it is periodic and many input values will map to the same output value. For example, \(\sin(0) = \sin(2\pi) = 0\), so it doesn't satisfy the one-to-one property for all real numbers.
04

Restrict the domain of the sine function

In order to satisfy the one-to-one property required for the sine function to have an inverse function, the domain of the sine function must be restricted. A common choice is to restrict the domain to the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\). In this interval, the sine function is one-to-one, as it is increasing and no two values of \(x\) will map to the same value of \(\sin(x)\).
05

Determine if the restricted sine function is onto over its range

Within the restricted domain of \([-\frac{\pi}{2}, \frac{\pi}{2}]\), the sine function takes on all values within the range of \([-1, 1]\) exactly once. Consequently, the restricted sine function is onto, and eventually satisfies both one-to-one and onto properties.
06

Conclusion

In order to define its inverse function, the domain of the sine function must be restricted, as it doesn't satisfy the one-to-one property when considering its whole domain. A common restricted domain is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), which ensures the sine function to be one-to-one and onto over its range, thus meeting the requirements for having an inverse function.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. \(\sin (a+b)=\sin a+\sin b\) b. The equation \(\cos \theta=2\) has multiple real solutions. c. The equation \(\sin \theta=\frac{1}{2}\) has exactly one solution. d. The function \(\sin (\pi x / 12)\) has a period of 12 e. Of the six basic trigonometric functions, only tangent and cotangent have a range of \((-\infty, \infty)\) f. \(\frac{\sin ^{-1} x}{\cos ^{-1} x}=\tan ^{-1} x\) g. \(\cos ^{-1}(\cos (15 \pi / 16))=15 \pi / 16\) h. \(\sin ^{-1} x=1 / \sin x\)

Use analytical methods to find the following points of intersection. Use a graphing utility only to check your work. Find the point(s) of intersection of the parabolas \(y=x^{2}\) and \(y=-x^{2}+8 x\)

Given the following information about one trigonometric function, evaluate the other five functions. $$\sec \theta=\frac{5}{3} \text { and } 3 \pi / 2<\theta<2 \pi$$

Use the following steps to prove that \(\log _{b}(x y)=\log _{b} x+\log _{b} y\). a. Let \(x=b^{p}\) and \(y=b^{q}\). Solve these expressions for \(p\) and \(q\) respectively. b. Use property El for exponents to express \(x y\) in terms of \(b, p\) and \(q\). c. Compute \(\log _{b}(x y)\) and simplify.

Determine whether the graphs of the following equations and functions have symmetry about the \(x\) -axis, the \(y\) -axis, or the origin. Check your work by graphing. $$x^{2 / 3}+y^{2 / 3}=1$$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.