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) = cos⁻¹ (x + 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
Multiple Choice
Given a right triangle where the length of the adjacent side to angle is and the hypotenuse is , in which triangle is the value of equal to (that is, )?
A
A triangle with an angle where the opposite side is and the hypotenuse is
B
A triangle with an angle where the opposite side is and the hypotenuse is
C
A triangle with an angle where the adjacent side is and the hypotenuse is
D
A triangle with an angle where the adjacent side is and the hypotenuse is
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Verified step by step guidance1
Recall the definition of cosine in a right triangle: for an angle \(x\), \(\cos(x) = \frac{\text{adjacent side}}{\text{hypotenuse}}\).
Given that \(x = \arccos\left(\frac{1}{2}\right)\), this means \(\cos(x) = \frac{1}{2}\).
Identify the triangle where the ratio of the adjacent side to the hypotenuse equals \(\frac{1}{2}\), which matches the given adjacent side length of 1 and hypotenuse length of 2.
Check the other options by calculating their cosine ratios: for example, if the opposite side is 1 and hypotenuse is 2, \(\cos(x) = \frac{\text{adjacent}}{2}\), which is not necessarily \(\frac{1}{2}\) without knowing the adjacent side; similarly, if the hypotenuse is smaller than the side, the ratio would be invalid.
Conclude that the triangle with adjacent side 1 and hypotenuse 2 corresponds to \(x = \arccos\left(\frac{1}{2}\right)\) because it satisfies the cosine ratio definition.
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