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Stitz-Zeager_College_Algebra_e-book

# The last quantity to nd is the phase unlike the

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Unformatted text preview: o π , π then tan(t) ≤ 0. As a 2 result, we get a piecewise deﬁned function for tan(t) √ tan(t) = x2 − 1, if 0 ≤ t < π 2 √ 2 − 1, if π < t ≤ π −x 2 Now we need to determine what these conditions on t mean for x. We know that the domain of arcsec(x) is (−∞, −1] ∪ [1, ∞), and since x = sec(t), x ≥ 1 corresponds to 0 ≤ t < π , and x ≤ −1 corresponds to π < t ≤ π . Since we encountered no further 2 2 restrictions on t, the equivalence below holds for all x in (−∞, −1] ∪ [1, ∞). √ tan(arcsec(x)) = x2 − 1, if x ≥ 1 √ − x2 − 1, if x ≤ −1 (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in π , 0 ∪ 0, π . Our objective is to write cos(arccsc(4x)) = cos(t) in terms of 2 2 x. Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 41 . The identity cos2 (t) + sin2 (t) = 1 holds x 2 for all values of t and substituting for sin(t) yields...
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