SECTION 7.1 The Inverse Sine, Cosine, and Tangent Functions 479 Solution (a) Let θ = − tan 1. 1 Then θ is the angle, π θ π − < < 2 2 , whose tangent equals 1. θ π θ π θ π θ π = − < < = − < < − tan 1 2 2 tan 1 2 2 1 Look at Table 3. The only angle θ within the interval π π ( ) − 2 , 2 whose tangent is 1 is π 4 , so π = − tan 1 4 1 (b) Let θ ( ) = − − tan 3 . 1 Then θ is the angle, π θ π − < < 2 2 , whose tangent equals − 3. θ π θ π θ π θ π ( ) = − − < < = − − < < − tan 3 2 2 tan 3 2 2 1 Look at Table 3.The only angle θ within the interval π π ( ) − 2 , 2 whose tangent is − 3 is π − 3 , so π ( ) − = − − tan 3 3 1 θ θ tan 2 π − Undefined 3 π − 3 − 4 π − 1− 6 π − 3 3 − 0 0 6 π 3 3 4 π 1 3 π 3 2 π Undefined Table 3 Now Work PROBLEM 15 7 Use Properties of Inverse Functions to Find Exact Values of Certain Composite Functions Recall from the discussion of functions and their inverses in Section 5.2 that ( ) ( ) = −f f x x 1 for all x in the domain of f and that ( ) ( ) = − f f x x 1 for all x in the domain of −f .1 In terms of the sine function and its inverse, these properties are of the form π π ( ) ( ) ( ) = = − ≤ ≤ − − f f x x x x sin sin where 2 2 1 1 (4a) ( ) ( ) ( ) = = − ≤ ≤ − − f f x x x x sin sin where 1 1 1 1 (4b) Finding the Exact Value of Certain Composite Functions Find the exact value, if any, of each composite function. (a) π ( ) − sin sin 8 1 (b) π ( ) − sin sin 5 8 1 EXAMPLE 7 Solution (a) The composite function π ( ) − sin sin 8 1 follows the form of property (4a). Because π 8 is in the interval π π ⎡ − ⎣ ⎢ ⎤ ⎦ ⎥ 2 , 2 , use property (4a) to conclude π π ( ) = − sin sin 8 8 1 Figure 14 verifies the calculation using a TI-84 Plus CE graphing calculator. Figure 14 (continued)
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