Textbook Question
In Exercises 39–54, find the exact value of each expression, if possible. Do not use a calculator. sin⁻¹ (sin 5π/6)
782
views
5. Inverse Trigonometric Functions and Basic Trigonometric Equations
Inverse Sine, Cosine, & Tangent
1:33 minutesProblem 49
Textbook Question
In Exercises 29–51, find the exact value of each expression. Do not use a calculator. sin⁻¹(sin π/3)
Verified step by step guidance1
Recognize that the function \( \sin^{-1}(x) \) (also called arcsin) is the inverse sine function, which returns an angle \( \theta \) such that \( \sin(\theta) = x \) and \( \theta \) lies within the principal range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
Identify the given expression: \( \sin^{-1}(\sin \frac{\pi}{3}) \). Here, \( \frac{\pi}{3} \) is the angle inside the sine function.
Note that \( \frac{\pi}{3} \) (which is 60 degrees) is not within the principal range of arcsin, since \( \frac{\pi}{3} > \frac{\pi}{2} \). Therefore, the value of \( \sin^{-1}(\sin \frac{\pi}{3}) \) will not simply be \( \frac{\pi}{3} \).
Use the property that for angles outside the principal range, \( \sin^{-1}(\sin x) = \pi - x \) if \( x \) is in the interval \( \left(\frac{\pi}{2}, \pi\right] \). Since \( \frac{\pi}{3} \) is less than \( \frac{\pi}{2} \), this property does not apply directly, so check the exact position of \( \frac{\pi}{3} \) relative to the principal range.
Since \( \frac{\pi}{3} \) is within \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), the arcsin function will return \( \frac{\pi}{3} \) itself. Therefore, \( \sin^{-1}(\sin \frac{\pi}{3}) = \frac{\pi}{3} \).
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
1mKey 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 sin⁻¹ or arcsin, returns the angle whose sine is a given value. Its output is restricted to the principal range of [-π/2, π/2] to ensure it is a function. Understanding this range is crucial when simplifying expressions involving sin⁻¹.
Recommended video:
4:03
Inverse Sine
Sine Function and Its Periodicity
The sine function is periodic with period 2π, meaning sin(θ) = sin(θ + 2kπ) for any integer k. Additionally, sine is symmetric about π/2, so sin(π - θ) = sin(θ). Recognizing these properties helps in finding equivalent angles within the principal range of arcsin.
Recommended video:
5:33Period of Sine and Cosine Functions
Evaluating sin⁻¹(sin θ) for Angles Outside the Principal Range
When θ is outside the principal range of arcsin, sin⁻¹(sin θ) equals the angle within [-π/2, π/2] that has the same sine value as θ. This often involves using angle identities or adjusting θ by subtracting or adding π to find the equivalent angle in the principal range.
Recommended video:
7:28Evaluate Composite Functions - Values Not on Unit Circle
Related Videos
Related Practice