In Exercises 29–51, find the exact value of each expression. Do not use a calculator. sec⁻¹ (−1)
Table of contents
- 0. Review of College Algebra4h 45m
- 1. Measuring Angles40m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
5. Inverse Trigonometric Functions and Basic Trigonometric Equations
Inverse Sine, Cosine, & Tangent
Problem 43
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)
Verified step by step guidance1
Recognize that the expression is \( \sin^{-1}(\sin(\frac{5\pi}{6})) \), which means we want to find the angle whose sine is \( \sin(\frac{5\pi}{6}) \).
Recall that \( \sin(\frac{5\pi}{6}) = \sin(\pi - \frac{\pi}{6}) = \sin(\frac{\pi}{6}) \), so \( \sin(\frac{5\pi}{6}) = \frac{1}{2} \).
Understand that the function \( \sin^{-1}(x) \) (also called arcsin) returns values in the principal range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
Since \( \sin^{-1}(\sin(\frac{5\pi}{6})) = \sin^{-1}(\frac{1}{2}) \), find the angle in the principal range whose sine is \( \frac{1}{2} \).
Identify that the angle in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) with sine \( \frac{1}{2} \) is \( \frac{\pi}{6} \), so \( \sin^{-1}(\sin(\frac{5\pi}{6})) = \frac{\pi}{6} \).
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
4mPlay a video:
0 Comments
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], meaning the result must lie within this interval to be considered the principal value.
Recommended video:
Inverse Sine
Sine Function Periodicity and Symmetry
The sine function is periodic with period 2π and is symmetric about the origin (odd function). This means sin(θ) = sin(π - θ), which helps in finding equivalent angles within the principal range when evaluating inverse sine expressions.
Recommended video:
Period 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 symmetry to find the equivalent angle inside the principal range.
Recommended video:
Evaluate Composite Functions - Values Not on Unit Circle
Related Videos
Related Practice
Textbook Question
717
views
