/*! 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 35 Is arctangent an even function, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 35

Is arctangent an even function, an odd function, or neither?

Short Answer

Expert verified
The arctangent function (tan^(-1)(x) or arctan(x)) is an odd function, as it satisfies the condition \(arctan(-x) = -arctan(x)\).

Step by step solution

01

Define the arctangent function and conditions for even and odd functions

The arctangent function is given by y = tan^(-1)(x) or y = arctan(x). Recall that for a function to be even, it must satisfy f(x) = f(-x), and for a function to be odd, it must satisfy f(-x) = -f(x).
02

Test if arctangent is an even function

To test if arctangent is an even function, we need to check if arctan(x) = arctan(-x). Let's try to find the relationship between these two: arctan(-x) can be rewritten as the angle whose tangent is -x. If we take the tangent of both sides, we get: tan(arctan(-x)) = -x We know that tangent is an odd function, which means tan(-y) = -tan(y). Thus, we can rewrite the left-hand side as: tan(-arctan(x)) = -x This equation tells us that the tangent of the angle -arctan(x) is equal to -x, but it does not imply that arctan(x) = arctan(-x). So, the arctangent function does not satisfy the condition for being even.
03

Test if arctangent is an odd function

Next, we'll test if arctangent is an odd function by checking if arctan(-x) = -arctan(x). From our work in Step 2, we have: tan(-arctan(x)) = -x Now, let's take the inverse tangent of both sides: -arctan(x) = arctan(-x) This condition holds true, which means arctan(x) is indeed an odd function.
04

Conclusion

After testing both the even and odd conditions, we find that arctangent function (tan^(-1)(x) or arctan(x)) is an odd function, since it satisfies the condition f(-x) = -f(x).

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

Find exact expressions for the indicated quantities. The following information will be useful: $$ \begin{array}{l} \cos 22.5^{\circ}=\frac{\sqrt{2+\sqrt{2}}}{2} \text { and } \sin 22.5^{\circ}=\frac{\sqrt{2-\sqrt{2}}}{2} \\ \cos 18^{\circ}=\sqrt{\frac{\sqrt{5}+5}{8}} \text { and } \sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \end{array} $$ [The value for \(\sin 22.5^{\circ}\) used here was derived in Example 4 in Section \(5.5 ;\) the other values were derived in Exercise 64 and Problems 102 and 103 in Section \(5.5 .]\) $$\sin 37.5^{\circ}$$

Find constants \(a, b,\) and \(c\) such that $$\sin ^{4} \theta=a+b \cos (2 \theta)+c \cos (4 \theta)$$ for all \(\theta\).

Find a formula for \(\cos (4 \theta)\) in terms of \(\cos \theta\).

Find constants \(a, b,\) and \(c\) such that $$\cos ^{4} \theta=a+b \cos (2 \theta)+c \cos (4 \theta)$$ for all \(\theta\).

Evaluate \(\sin \left(\cos ^{-1} \frac{1}{4}+\tan ^{-1} 2\right)\).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

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.