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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