Stitz-Zeager_College_Algebra_e-book

4754 chapter 11 applications of trigonometry 111

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Unformatted text preview: t) = −5 by definition, we have cot(arccot(−5)) = −5. 3 (d) We start simplifying sin arctan − 4 by letting t = arctan − 3 . Then t lies between 4 3 − π and π with tan(t) = − 4 . We seek sin arctan − 3 = sin(t). There are many 2 2 4 ways to proceed at this point. The Pythagorean Identity, 1 + cot2 (t) = csc2 (t) relates the reciprocals of sin(t) and tan(t), so this seems a reasonable place to start. Since 5 1 42 tan(t) = − 3 , cot(t) = tan(t) = − 4 . We get 1 + − 3 = csc2 (t) so that csc(t) = ± 3 , 4 3 3 and, hence, sin(t) = ± 3 . Since − π < t < π and tan(t) = − 4 < 0, it must be the case 5 2 2 3 that t lies between − π and 0. As a result, we choose sin(t) = − 5 . 2 2. (a) If we let t = arctan(x), then − π < t < π and tan(t) = x. We look for a way to express 2 2 tan(2 arctan(x)) = tan(2t) in terms of x. Before we get started using identities, we note that tan(2t) is undefined when 2t = π + πk for integers k , which means we need to 2 exclude any of the values t = π + π k , where k is an integer, which lie in − π , π . We 4 2 22 find that we need to discard t = ± π from the discussion, so we are now working with 4 t in − π ,...
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