Stitz-Zeager_College_Algebra_e-book

The last quantity to nd is the phase unlike the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online