Textbook Question
Find the exact value of each real number y if it exists. Do not use a calculator.
y = sin⁻¹ (―1)
Ch. 6 - Inverse Circular Functions and Trigonometric Equations
Lial12th EditionTrigonometryISBN: 9780136552161
Not the one you use?Change textbookSelect a different textbook
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
All textbooks
Lial 12th EditionCh. 6 - Inverse Circular Functions and Trigonometric EquationsProblem 13
Lial 12th EditionCh. 6 - Inverse Circular Functions and Trigonometric EquationsProblem 13Chapter 7, Problem 13
Find the exact value of each real number y if it exists. Do not use a calculator.
y = sin⁻¹ 0
Verified step by step guidance1
Understand that the problem asks for the values of \(y\) such that \(\sin y = 0\), where \(y = \sin^{-1} 0\) represents the inverse sine (arcsine) function.
Recall the definition of the inverse sine function: \(y = \sin^{-1} x\) means \(\sin y = x\) and \(y\) lies within the principal range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
Set up the equation based on the problem: \(\sin y = 0\) with \(y \in [-\frac{\pi}{2}, \frac{\pi}{2}]\).
Identify all angles \(y\) in the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\) where \(\sin y = 0\). These are the values where the sine function crosses zero within the principal range.
Conclude that the exact value(s) of \(y\) satisfying the equation are those angles found in the previous step, which represent the output of \(\sin^{-1} 0\).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inverse Sine Function (sin⁻¹ or arcsin)
The inverse sine function, denoted as sin⁻¹ or arcsin, returns the angle whose sine is a given number. It is defined for inputs between -1 and 1 and outputs angles in the range [-π/2, π/2]. Understanding this function helps find angles corresponding to specific sine values.
Recommended video:
4:03
Inverse Sine
Sine Function Values
The sine function relates an angle to the ratio of the opposite side over the hypotenuse in a right triangle. Key values include sin(0) = 0, sin(π/2) = 1, and sin(-π/2) = -1. Recognizing these standard values aids in identifying angles for given sine outputs.
Recommended video:
5:08Sine, Cosine, & Tangent of 30°, 45°, & 60°
Domain and Range Restrictions of Inverse Trigonometric Functions
Inverse trigonometric functions have restricted ranges to ensure they are functions. For arcsin, the output is limited to angles between -π/2 and π/2. This restriction means that for sin⁻¹(0), the principal value is 0, even though sine of other angles like π also equals zero.
Recommended video:
4:22Domain and Range of Function Transformations
Related Practice
Textbook Question
Find the exact value of each real number y. Do not use a calculator.
y = arccot (―1)
1
views
Textbook Question
Find the exact value of each real number y. Do not use a calculator.
y = sec⁻¹ (―2)
Textbook Question
Use the unit circle shown here to solve each simple trigonometric equation. If the variable is x, then solve over [0, 2π). If the variable is θ, then solve over [0°, 360°).
<IMAGE>
sin θ = ―√2/2
Textbook Question
Solve each equation for x, where x is restricted to the given interval.
y = ― 2 cos 5x , for x in [0, π/5]
Textbook Question
Solve each equation for x, where x is restricted to the given interval.
y = 6 cos x/4 , for x in [0, 4π]