Stitz-Zeager_College_Algebra_e-book

We rst go after the vertical shift b whose value

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Unformatted text preview: yields arcsec sec 54 = arcsec(− 2) = arccos − 22 = 34 . (d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). Then, csc(t) = −3 and, since this is negative, we have that t lies in the interval − π , 0 . We are after 2 cot (arccsc (−3)) = cot(t), so we use the Pythagorean √ Identity √ + cot2 (t) = csc2 (t). 1 Substituting, we have 1 + cot2 (t) = (−3)2 , or cot(t) = ± 8 = ±2 2. Since − π ≤ t < 0, 2 √ cot(t) < 0, so we get cot (arccsc (−3)) = −2 2. 2. (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π ∪ π , π , and we seek a formula for tan(t). Since tan(t) is defined for all 2 2 the t values in 0, π ∪ π , π , we have no additional restrictions on t. The identity 2 2 1 + tan2 (t) = sec2 (t) is valid for all values t under consideration, and we get substitute √ sec(t) = x to get 1 + tan2 (t) = x2 . Hence, tan(t) = ± x2 − 1. If t belongs to 0, π 2 then tan(t) ≥ 0; if, on the the other hand, t belongs t...
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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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