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Chapter 5: Problem 82

Verifying Identities In Exercises 81 and \(82,\) verify each identity. \((a) \arcsin (-x)=-\arcsin x, \quad|x| \leq 1\) \((b) \arccos (-x)=\pi-\arccos x, \quad|x| \leq 1\)

Short Answer

Expert verified
We verified the identities \(\arcsin(-x)=-\arcsin x\) and \(\arccos(-x)=\pi-\arccos x\). This was done by considering the properties and definitions of the arc functions.

Step by step solution

01

Understand the Problem

We have been given two identities that involve the arcsin and arccos functions, respectively. The goal is to verify these identities. For each, we need to show that the left-hand side equals the right-hand side. We will make use of the properties of the arc functions.
02

Verify (a) Identity

Consider \(y=\arcsin(-x)\). Solving for x, \(x=-\sin y\). The domain of arcsin is \(-1\le x\le 1\), hence \(-1 \le -\sin y \le 1\), which implies \(-1 \le \sin y \le 1\). Therefore, y is in fact sin's range \(-\pi / 2 \le y \le \pi / 2\) or \(\pi / 2 \le -y \le \pi / 2\). This means y = - arcsin(x). Hence, \(\arcsin(-x)=-\arcsin x\).
03

Verify (b) Identity

Consider \(y=\arccos(-x)\). Solving for x, \(x=-\cos y\). The domain of arccos is \(-1\le x\le 1\), hence \(-1 \le -\cos y \le 1\), which implies \(-1 \le \cos y \le 1\). Therefore, y is in fact cos's range \(0 \le y \le \pi\) or \(0 \le \pi - y \le \pi\). This means y = \(\pi - \arccos x\). Hence, \(\arccos(-x)=\pi-\arccos x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arcsin Function
The arcsin function, also known as the inverse sine function, is represented as \(\arcsin(x)\). This function is used to find the angle whose sine is \(x\).
Some important things to know about the arcsin function:
  • Its domain is limited to the interval \([-1, 1]\).
  • The range of the arcsin function is \([ -\frac{\pi}{2}, \frac{\pi}{2} ]\).
  • \(\arcsin(x)\) is the angle \(y\) such that \(\sin(y) = x\), and \(-\frac{\pi}{2} \le y \le \frac{\pi}{2}\).
A useful property of arcsin is that \(\arcsin(-x) = -\arcsin(x)\). This symmetric property makes calculations easier when negative inputs are involved. It tells you that if you take the sine of a negative number, the resulting angle will simply be the negative of the angle for the positive number.
This property helps greatly in trigonometric problems, as seen when proving identities.
Arccos Function
The arccos function is the inverse of the cosine function and is denoted as \(\arccos(x)\). This function helps in determining the angle whose cosine is \(x\).
Key points about the arccos function include:
  • Its domain is within \([-1, 1]\).
  • The range is \([0, \pi]\).
  • \(\arccos(x)\) is the angle \(y\) such that \(\cos(y) = x\), and \(0 \le y \le \pi\).
The property \(\arccos(-x) = \pi - \arccos(x)\) is quite useful, especially when dealing with negative \(x\).
It indicates that when you flip the sign of the input \(x\), the found angle will differ from \(\pi\) by the positive angle.
This understanding helps in solving trigonometric equations where angle transformations are necessary.
Function Properties
Trigonometric functions, like arcsin and arccos, have specific properties that are beneficial in problem-solving and verifying identities.
For the arcsin function:
  • Symmetry: \(\arcsin(-x) = -\arcsin(x)\)
  • Monotonicity: The arcsin function is strictly increasing.
For the arccos function:
  • Symmetry: \(\arccos(-x) = \pi - \arccos(x)\)
  • Monotonicity: The arccos function is strictly decreasing.
These properties allow you to predict behavior, simplify calculations, and confirm inverse function characteristics in trigonometry.
By recognizing these, we not only simplify our processes but ensure accuracy in finding angles and verifying identities.
Verifying Identities
Verifying trigonometric identities involves showing that two expressions are equal for all values in their domain. It requires a mix of algebraic manipulation and knowledge of trigonometric properties.
Here is how you can effectively verify identities:
  • Substitute known values and observe the results.
  • Use algebraic skills to manipulate and transform expressions.
  • Apply trigonometric properties to simplify expressions.
In our exercise, knowing properties like \(\arcsin(-x) = -\arcsin(x)\) and \(\arccos(-x) = \pi - \arccos(x)\) were pivotal.
These properties directly helped demonstrate the truth of each identity by transforming one side of the equation to match the other.
This process not only tests your understanding of trigonometric concepts but also strengthens your problem-solving abilities.

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Most popular questions from this chapter

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result of part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$\frac{d v}{d t}=-\left(32+k v^{2}\right)$$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part \((c)\) for \(k=0.001 .\) Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral $$\int_{0}^{t_{0}} v(t) d t$$ where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

In Exercises 47-50, find the indefinite integrals, if possible, using the formulas and techniques you have studied so far in the text. $$\begin{array}{l}{\text { (a) } \int \sqrt{x-1} d x} \\ {\text { (b) } \int x \sqrt{x-1} d x} \\ {\text { (c) } \int \frac{x}{\sqrt{x-1}} d x}\end{array}$$

Proof \(\quad\) Graph \(y_{1}=\frac{x}{1+x^{2}}, y_{2}=\arctan x,\) and \(y_{3}=x\) on \([0,10] .\) Prove that \(\frac{x}{1+x^{2}} <\arctan x< x\) for \(x >0\)

Using the Area of a Region Find the value of \(a\) such that the area bounded by \(y=e^{-x},\) the \(x\) -axis, \(x=-a,\) and \(x=a\) is \(\frac{8}{3}\)

Evaluate \(\lim _{x \rightarrow \infty}\left[\frac{1}{x} \cdot \frac{a^{x}-1}{a-1}\right]^{1 / x}\) where \(a>0, a \neq 1\)
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