The graphs of y = sin⁻¹ x, y = cos⁻¹ x, and y = tan⁻¹ x are shown in Table 2.8. In Exercises 97–106, use transformations (vertical shifts, horizontal shifts, reflections, stretching, or shrinking) of these graphs to graph each function. Then use interval notation to give the function's domain and range. f(x) = sin⁻¹ x + π/2
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
Multiple Choice
For which interval of is the cancellation property valid?
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D
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Verified step by step guidance1
Recall that the function \( \sin^{-1}(\sin(x)) = x \) holds true only when \( x \) is within the principal domain of the inverse sine function, which is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
Understand that the inverse sine function, \( \sin^{-1}(y) \), is defined to return values only in the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) to ensure it is a proper function (one-to-one).
Since \( \sin(x) \) is periodic and not one-to-one over all real numbers, outside the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), the equality \( \sin^{-1}(\sin(x)) = x \) does not hold because the inverse sine will return the principal value, not \( x \) itself.
Therefore, to find the interval where the cancellation property holds, identify the domain of \( x \) such that \( x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
Conclude that the cancellation property \( \sin^{-1}(\sin(x)) = x \) is valid only for \( x \) in the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
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