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4.1.20001 - i x E c E i r E E a E'E i t a E E 5 é a 3 $3...

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Unformatted text preview: i x E c E i r E E a E 'E: i; § % t a; E E 5 é a 3 $3 212 Chapter 4 Solving Conditional Trigonometric Equations Figure 4.2 EXAMPLE 1 Evaluating the inverse sine function Note that the nth power of the sine function is usually written as sin”(x§ shorthand notation for (sin x)", provided It ¢ —1. The —l used in sin‘1 x indi the 1nverse function and does not mean reciprocal. To write 1 / s1n x using exponen’ we must write (sin x).1 ,9 Find the exact value of each expression without using a table or a calculator. a. sin—1(l/2) b. arcsin(—\/§/2) Solution . a. The value of SID—1(1/2) 15 the number a in the interval [*7T/27T/2]kS11Ch that sin(a ) — 1/2. We recall that sin(7T/6)— — 1/2, and so sin—1(l/2) — 77/6. Note that 77/6 1s the only value ofoz 1n [— 77/2, 77/2] for which sin(a)— — l/2. b. The value of arcsin(— \/_/2) is the number a in the interval[ [—7T/2, 77/2]? such that sin(a) = —\/§/2. Since sin(—7r/3) = —\/§/2, we hav . ’ arcsin(-\/§/2) = —77/3. Note that —’7T/3 is the only value ofa in [—7r/2, 7T/2I for which sin(a) = —\/§/2. . r >TRY THIS. Find the exact value ofsin*1(— V22). , ,- _' EXAMPLE 2 Evaluating the inverse sine function Find the exact value of each expression in degrees without using a table or a calculator. Ii 3. sin“1 ( V2/ 2) b. arcsin(0) Solution a. The value of sin_1(\/_/ 2) in degrees is the angle at in the interval [- 90°, 90°] ' such that sin(a)= V2/2. We recall that sin(45°)— — V2/2, and so sin‘1(\/2/2)— — 45°. ’ b. The value of arcsin(0 ) 1n degrees 15 the angle a in the interval [— 90°, 90°] for . which sin(a) = 0. Since sin(0°)— — 0, we havearcsin(0)— — 70°. >TRY THIS. Find the exact value in degrees of arcsin( — l / 2). , I , ‘ In the next example we use a calculator to find the degree measure of an angle whose sine is given. To obtain degree measure make sure the calculator is in degree mode. Scientific calculators usually have a key labeled sin“1 that gives values for the inverse sine function. EXAMPLE 3 Finding an angle given its sine Suppose that a is an angle such that —90° < at < 90°. In each case, find at to the nearest tenth of a degree; a. sina = 0.88 b. sina = —0.27 Solution a. The value of sin—1(0.88) is the only angle in [—90°, 90°] with a sine of 0.88. Use a calculator in degree mode to get a = sin—1(0.88) % 61.6°. b. Use a calculator to get a = sin—1(—0.27) % —15.7°. E Figure 4.2 shows how to find the angle in part (a) on a graphing calculator and how to check. Make sure that the mode is degrees. VTRY THIS. Find a to the nearest tenth of a degree if —90° < a < 90° and sin a =0. 257. I ...
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