Stitz-Zeager_College_Algebra_e-book

As far as the initial velocity v0 is concerned v0 0

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Unformatted text preview: for all t under consideration, we have no additional restrictions on t. The identity 1 + tan2 (t) = sec2 (t) is valid for all t under discussion, so we substitute sec(t) = x to get 1 + tan2 (t) = √ π x2 . We get tan(t) = ± x2 − 1, but since t lies in 0, π ∪ π , 32 , tan(t) ≥ 0, so we 2 √ x choose tan(t) = √ 2 − 1. Since we found no additional restrictions on t, the equivalence tan(arcsec(x)) = x2 − 1 holds on the domain of arcsec(x), (−∞, −1] ∪ [1, ∞). π (b) If we let t = arccsc(4x), then csc(t) = 4x for t in 0, π ∪ π , 32 . Then cos(arccsc(4x)) = 2 cos(t) and our objective is to express the latter in terms of x. Since cos(t) is deﬁned everywhere, we have no additional restrictions on t. From csc(t) = 4x, we have sin(t) = 1 1 2 2 csc(t) = 4x which allows us to use the Pythagorean Identity, cos (t) + sin (t) = 1, which holds for all values of t. We get cos2 (t) + 12 4x 16x2 −1 16x2 = 1, or cos(t) = ± √ =± 16x2 −1 . 4|x| π If t lies in 0, π , then cos(t) ≥ 0. Otherwis...
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