732 CHAPTER 7 Trigonometric Identities and Equations EXAMPLE 4 Finding Inverse Function Values with a Calculator Use a calculator to approximate each value. (a) Find y in radians if y = csc-11-32. (b) Find u in degrees if u = arccot1-0.35412. SOLUTION (a) With the calculator in radian mode, enter csc-1 1-32 as sin-1 A 1 -3B to obtain y ≈ -0.3398369095. See Figure 22(a). (b) A calculator in degree mode gives the inverse tangent value of a negative number as a quadrant IV angle. The restriction on the range of arccotangent implies that u must be in quadrant II. arccot1-0.35412 is entered as tan-1 A 1 -0.3541B + 180°. As shown in Figure 22(b), u ≈109.4990544°. S Now Try Exercises 53 and 65. (a) (b) Figure 22 CAUTION Be careful when using a calculator to evaluate the inverse cotangent of a negative quantity. Enter the inverse tangent of the reciprocal of the negative quantity, which returns an angle in quadrant IV. Because inverse cotangent is negative in quadrant II, adjust the calculator result by adding p or 180° accordingly. (Note that cot-1 0 = p 2 or 90°.) EXAMPLE 5 Finding Function Values Using Definitions of theTrigonometric Functions Evaluate each expression without using a calculator. (a) sin atan-1 3 2b (b) tan acos-1 a- 5 13bb SOLUTION (a) Let u = tan-1 3 2, and thus tan u = 3 2 . The inverse tangent function yields values only in quadrants I and IV, and because 3 2 is positive, u is in quadrant I. Sketch u in quadrant I, and label a triangle, as shown in Figure 23 on the next page. By the Pythagorean theorem, the hypotenuse is 213. The value of sine is the quotient of the side opposite and the hypotenuse. sin atan-1 3 2b = sin u = 32 13 = 32 13 # 213 213 = 3213 13 Rationalize the denominator. Use the following to evaluate these inverse functions on a calculator. sec−1 x is evaluated as cos−1 1 x ; csc−1 x is evaluated as sin−1 1 x ; cot−1 x is evaluated as e tan−1 1 x if x +0 180° +tan−1 1 x if x *0. Degree mode
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