Stitz-Zeager_College_Algebra_e-book

Using extended interval notation we 4k 1 k 1 4 742

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Unformatted text preview: an(t) in terms of x. This is where identities come into play, but we must be careful to use identities which are defined for all values of t under consideration. In this situation, we have 0 ≤ t ≤ π , but since the quantity we are looking for, tan(t), is undefined at t = π , the identities we choose to need 2 sin( to hold for all t in 0, π ∪ π , π . Since tan(t) = cos(t) , and we know cos(t) = x, all that 2 2 t) remains is to find sin(t) in terms of x and we’ll be done.4 The identity cos2 (t)+sin2 (t) = 1 holds for all t, in particular the ones in 0, π ∪ π , π , so substituting cos(t) = x, we 2 √2 get x2 + sin2 (t) = 1. Hence, sin(t) = ± 1 − x2 and since t belongs to 0, π ∪ π , π , 2 2 √ √ sin( −2 sin(t) ≥ 0, so we choose sin(t) = 1 − x2 . Thus, tan(t) = cos(t) = 1x x . To determine t) the values of x for which this is valid, we first note that arccos(x) is valid only for −1 ≤ x ≤ 1. Additionally, as we have already mentioned, tan(t) is not defined √ when −2 π π t = 2 , which means we must exclude x = cos 2 = 0. Hence, tan (arccos (x)) = 1x x for x in [−1, 0) ∪ (0, 1]. (b) We proceed as in the previous problem by writing 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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