4.1.60001 - 216 Chapter 4 Solving Conditional Trigonometric...

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Unformatted text preview: 216 Chapter 4 Solving Conditional Trigonometric Equations Domain (-oo, -1] U [1. 00) Range Pg, 0) L406] Identities for 030“, sec", and cot‘1 Domain (-00,-1]U[1,00) ‘ Domain (—00,00) Range [OH-72!)U(§77] Range (0,77) When studying inverse trigonometric functions, you should first learn to ev ‘ ate sin’I, cosfil, and tan—1. Those values can be used along with the ident csc a = l/sin oz, see a = l/cos a, and cota = 1/tanoz to evaluate csc_1, sec and cotgl. For example, sin(77/6) = 1/2 and csc(77/6) = 2. So the angle whose}; cosecant is 2 is the same as the angle whose sine is l / 2. In symbols, 1 77 —1 = ' —1 — = — csc (2) sm < 2) 6 . 1 111 general, 080—196 =Vsin_1(1/x). Likewise, sec’ x = cos’1(1/x). For the inverse: cotangent, cot—ix = tan—1(1/x) only for positive values of x, because of the choi of (0, 77) as the range of the inverse cotangent. We have cot—1(0) = 77/2 and if x is; negative cot—1(x) = tan’1(l/x) + 77. i Another relationship between cot_1 and tan_1 can be obtained from their graphs : in the Function Gallery on pages 215—216. Since the graph of y = cot—1(x)rcan be ob- :I tained from the graph of y = tan—1(x) by reflecting with respect to the x-axis and? translating upward a distance of 77/2, we have cot—1(x) = i—tan_1(x) + 7/2 or (x) = 77/2 — tan“1(x). These identities are summarized below. g) V /x)for|x|‘: .1, ;_ tan—1(l/x) _ i ’- fory)x«>,; ,0 , i: tanfl(’l/x)-l- n77 I‘forjé <70 ,_77/2,:' T forx:0 _‘ - EXAMPLE 7 Evaluating the inverse functions Find the exact value of each expression without using a table or a calculator. a. arcsec(—2) b. csc‘1(\/2) c. arccot(—1/\/§) Solution 3. To evaluate the inverse secant we use the identity sec—1(x) = cos—1(1/x). In this case, the arc whose secant is —2 is the same as the arc whose cosine is — 1 / 2. So we must find cos—1(—-1/2). Since cos(277/3) = —l/2 and since 277/3 is in the range of arccos, we have arccos(—1/2) = 277/3. So 2 arcsec(—2) = arccos(—1/2) = 771-. ...
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