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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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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