Stitz-Zeager_College_Algebra_e-book

this is a consequence of newtons second law of

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Unformatted text preview: into an arcsine problem. Instead, we appeal to the definition. π The real number t = arccsc(−2) lies between 0 and π or between π and 32 and satisfies 2 7π 7π csc(t) = −2. We have t = 6 , so arccsc(−2) = 6 . π π (c) Since 54 lies between π and 32 , we may apply Theorem 10.29 directly to simplify π π = 54 . We encourage the reader to work this through using the definiarcsec sec 54 tion as we have done in the previous examples to see how it goes. (d) To simplify cot (arccsc (−3)) we let t = arccsc (−3) so that cot (arccsc (−3)) = cot(t). π We know csc(t) = −3, and since this is negative, t lies between π and 32 . Using the Pythagorean Identity 1 + cot2 (t) = csc2 (t), we find 1 + cot2 (t) = (−3)2 so that √ √ π cot(t) = ± 8 = ±2 2. Since t is in the interval 0, 32 , we know cot(t) > 0. Our √ answer is cot (arccsc (−3)) = 2 2. π 2. (a) To simplify tan(arcsec(x)), we let t = arcsec(x) so sec(t) = x for t in 0, π ∪ π , 32 . Our 2 goal is to express tan(arcsec(x)) = tan(t) in terms of x. Since tan(t) is defined...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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