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4.7 INVERSE TRIGONOMETRIC FUNCTIONS Copyright © Cengage Learning. All rights reserved. What You Should Learn • Evaluate and graph the inverse sine function. • Evaluate and graph the other inverse trigonometric functions. • Evaluate and graph the compositions of trigonometric functions. 2 Inverse Sine Function 3 Inverse Sine Function For a function to have an inverse function, it must be one-to-one—that is, it must pass the Horizontal Line Test. Restrict the domain to the interval – / 2 x / 2, the following properties hold. 1. On the interval [– / 2, / 2], the function y = sin x is increasing. sin x has an inverse function on this interval. By definition, the values of inverse trigonometric functions are always in radians. 2. On the interval [– / 2, / 2], y = sin x takes on its full range of values, –1 sin x 1. 3. On the interval [– / 2, / 2], y = sin x is one-to-one. 4 Inverse Sine Function On the restricted domain – / 2 x / 2, y = sin x has a unique inverse function called the inverse sine function. y = arcsin x or y = sin –1 x. means the angle (or arc) whose sine is x. 5 Inverse Sine Function 6 Example 1 – Evaluating the Inverse Sine Function If possible, find the exact value. a. Solution: a. Because b. c. for . , Angle whose sine is 7 Example 1 – Solution b. Because for . cont’d , Angle whose sine is c. It is not possible to evaluate y = sin –1 x when x = 2 because there is no angle whose sine is 2. Remember that the domain of the inverse sine function is [–1, 1]. 8 Example 2 – Graphing the Arcsine Function Sketch a graph of y = arcsin x. Solution: In the interval [– / 2, / 2], 9 Example 2 – Solution cont’d y = arcsin x Domain: [–1, 1] Range: [– / 2, / 2] 10 Other Inverse Trigonometric Functions 11 Other Inverse Trigonometric Functions The cosine function is decreasing and one-to-one on the interval 0 x . cos x has an inverse function on this interval. On the interval 0 x the cosine function has an inverse function—the inverse cosine function: y = arccos x or y = cos –1 x. 12 Other Inverse Trigonometric Functions DOMAIN: [–1,1] RANGE: DOMAIN: [–1,1] RANGE: [0, ] DOMAIN: RANGE: 13 Example 3 – Evaluating Inverse Trigonometric Functions Find the exact value. a. arccos b. cos –1(–1) c. arctan 0 d. tan –1 (–1) Solution: a. Because cos ( / 4) = , and / 4 lies in [0, ], . Angle whose cosine is 14 Example 3 – Solution cont’d b. Because cos = –1, and lies in [0, ], cos –1(–1) = . Angle whose cosine is –1 c. Because tan 0 = 0, and 0 lies in (– / 2, / 2), arctan 0 = 0. Angle whose tangent is 0 15 Example 3 – Solution cont’d d. Because tan(– / 4) = –1, and – / 4 lies in (– / 2, / 2), tan –1 (–1) = . Angle whose tangent is –1 16 Compositions of Functions 17 Compositions of Functions For all x in the domains of f and f –1, inverse functions have the properties f (f –1(x)) = x and f –1(f (x)) = x. 18 Compositions of Functions These inverse properties do not apply for arbitrary values of x and y. . The property arcsin(sin y) = y is not valid for values of y outside the interval [– / 2, / 2]. 19 Example 5 – Using Inverse Properties If possible, find the exact value. a. tan[arctan(–5)] b. c. cos(cos –1 ) Solution: a. Because –5 lies in the domain of the arctan function, the inverse property applies, and you have tan[arctan(–5)] = –5. 20 Example 5 – Solution cont’d b. In this case, 5 / 3 does not lie within the range of the arcsine function, – / 2 y / 2. However, 5 / 3 is coterminal with which does lie in the range of the arcsine function, and you have 21 Example 5 – Solution cont’d c. The expression cos(cos –1) is not defined because cos –1 is not defined. Remember that the domain of the inverse cosine function is [–1, 1]. 22 Example 6 – Evaluating Compositions of Functions Find the exact value: tan 2 arccos 3 2 arccos , 3 2 3 Let 𝑢 = then cos 𝑢 = > 0, therefore, u is in QI 2 sin 𝑢 tan arccos = tan 𝑢 = = 3 cos 𝑢 2 1− 3 2 3 2 5 = 2 23 Example 6 – Evaluating Compositions of Functions Find the exact value: cos Let 𝑢 = Q IV 3 arc𝑠𝑖𝑛(− ), 5 3 cos arc𝑠𝑖𝑛(− ) = 5 3 arc𝑠𝑖𝑛(− ) 5 then sin 𝑢 = 3 1− − 5 3 − 5 2 < 0, therefore, u is in 4 = 5 24 Example Write the expression as an algebraic expression in x: 1 sin arccos 3𝑥 , 0 ≤ 𝑥 ≤ 3 sin arccos 3𝑥 = 1 − 9𝑥 2 25