Textbook Question
Find the exact value of each real number y. Do not use a calculator.
y = sin⁻¹ √2/2
Ch. 6 - Inverse Circular Functions and Trigonometric Equations
Lial12th EditionTrigonometryISBN: 9780136552161
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Table of contents
- Ch. R - Algebra Review381
- Ch. 1 - Trigonometric Functions188
- Ch. 2 - Acute Angles and Right Triangles168
- Ch. 3 - Radian Measure and The Unit Circle155
- Ch. 4 - Graphs of the Circular Functions100
- Ch. 5 - Trigonometric Identities238
- Ch. 6 - Inverse Circular Functions and Trigonometric Equations158
- Ch. 7 - Applications of Trigonometry and Vectors148
- Ch. 8 - Complex Numbers, Polar Equations, and Parametric Equations15
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Lial 12th EditionCh. 6 - Inverse Circular Functions and Trigonometric EquationsProblem 5
Lial 12th EditionCh. 6 - Inverse Circular Functions and Trigonometric EquationsProblem 5Chapter 7, Problem 5
Which one of the following equations has solution π?
a. arccos (―1) = x
b. arccos 1 = x
c. arcsin (―1) = x
Verified step by step guidance1
Recall the definition of the inverse cosine function, arccos, which returns an angle \(x\) in the range \(0 \leq x \leq \pi\) such that \(\cos x\) equals the given value.
Evaluate each option by considering the cosine or sine values at \(x = \pi\):
For option (a), check if \(\arccos(-1) = \pi\) because \(\cos \pi = -1\); this suggests \(x = \pi\) is a solution for (a).
For option (b), \(\arccos(1) = 0\) since \(\cos 0 = 1\), so \(x = \pi\) is not a solution here.
For option (c), \(\arcsin(-1) = -\frac{\pi}{2}\) because \(\sin(-\frac{\pi}{2}) = -1\), so \(x = \pi\) is not a solution here.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inverse Trigonometric Functions
Inverse trigonometric functions, such as arccos and arcsin, return the angle whose trigonometric value matches the given input. They are used to find angles from known sine or cosine values, with outputs restricted to specific principal value ranges.
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Introduction to Inverse Trig Functions
Range and Domain of arccos and arcsin
The arccos function has a domain of [-1, 1] and a range of [0, π], meaning it outputs angles between 0 and π radians. The arcsin function also has a domain of [-1, 1] but a range of [-π/2, π/2], so it outputs angles between -π/2 and π/2 radians.
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Evaluating arccos and arcsin at Specific Values
Evaluating arccos(-1) yields π because cos(π) = -1, arccos(1) yields 0 since cos(0) = 1, and arcsin(-1) yields -π/2 because sin(-π/2) = -1. Understanding these specific values helps identify which equation has π as a solution.
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Related Practice
Textbook Question
The point (π/4, 1) lies on the graph of y = tan x. Therefore, the point _______ lies on the graph of y = tan⁻¹ x.
Textbook Question
Decide whether each statement is true or false. If false, explain why.
The tangent and secant functions are undefined for the same values.
Textbook Question
The following equations cannot be solved by algebraic methods. Use a graphing calculator to find all solutions over the interval [0, 2π). Express solutions to four decimal places.
x² + sin x - x³ - cos x = 0
Textbook Question
Solve each equation for all exact solutions, in degrees.
2√3 cos (θ/2) = -3
Textbook Question
Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.
√2 cos 2θ = -1