2. Consider the function f: R → [-1, 1] defined by f(x) = sin(x) and the function g: [-π/2, π/2] → [-1, 1] defined by g(x) = sin(x). A. The function f is not injective, and the function g is injective. Explain. B. The inverse function of g will have a domain of [-1, 1] and a range of [-T/2, π/2]. Explain. (Henceforth, we will refer to the inverse function to g as y = sin ¹(x).) C. For any on the interval [-1, 1], the value of cos (sin-¹(x)) will be nonnegative. Explain. D. Utilizing the identity sin² (y) + cos² (y) = 1 and your result from Part C above, deduce that cos(sin-¹(x)) is equal to √1-x²). - x² (and not
2. Consider the function f: R → [-1, 1] defined by f(x) = sin(x) and the function g: [-π/2, π/2] → [-1, 1] defined by g(x) = sin(x). A. The function f is not injective, and the function g is injective. Explain. B. The inverse function of g will have a domain of [-1, 1] and a range of [-T/2, π/2]. Explain. (Henceforth, we will refer to the inverse function to g as y = sin ¹(x).) C. For any on the interval [-1, 1], the value of cos (sin-¹(x)) will be nonnegative. Explain. D. Utilizing the identity sin² (y) + cos² (y) = 1 and your result from Part C above, deduce that cos(sin-¹(x)) is equal to √1-x²). - x² (and not
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.1: Inverse Functions
Problem 55E
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