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Ch. 2 - Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Not the one you use?Change textbook
Chapter 2, Problem 49

In Exercises 29–51, find the exact value of each expression. Do not use a calculator. sin⁻¹(sin π/3)

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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, \( \sin^{-1}(\sin \frac{\pi}{3}) = \frac{\pi}{3} \).

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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 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.
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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.
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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.
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