Stitz-Zeager_College_Algebra_e-book

Next we recall that for the square root to be dened

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Unformatted text preview: 6 3 2. We find t = π , so that arccos cos 6 π 6 = π. 6 does not fall between 0 and π , Theorem 10.26 does not apply. We are forced to work through from the inside out starting with arccos cos √ the previous problem, we know arccos 3 2 11π 6 √ = arccos = π . Hence, arccos cos 6 11π 6 3 2 . From = π. 6 (g) To help simplify cos arccos − 3 let t = arccos − 3 . Then, by definition, 0 ≤ t ≤ π 5 5 and cos(t) = − 3 . Hence, cos arccos − 3 = cos(t) = − 3 . 5 5 5 704 Foundations of Trigonometry (h) As in the previous example, we let t = arccos − 3 so that 0 ≤ t ≤ π and cos(t) = − 3 . 5 5 In terms of t, then, we need to find sin arccos − 3 = sin(t). Using the Pythagorean 5 2 4 Identity cos2 (t)+sin2 (t) = 1, we get − 3 +sin2 (t) = 1 or sin(t) = ± 5 . Since 0 ≤ t ≤ π , 5 4 3 4 we choose3 sin(t) = 5 . Hence, sin arccos − 5 = 5 . 2. (a) We begin rewriting tan (arccos (x)) using t = arccos(x). We know that 0 ≤ t ≤ π and cos(t) = x, so our goal is to express tan (arccos (x)) = t...
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