In Exercises 52–53, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. sec(sin⁻¹ 1/x)
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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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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