Stitz-Zeager_College_Algebra_e-book

# When does the object rst pass through the equilibrium

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Unformatted text preview: e, t belongs to π , 32 in which case cos(t) ≤ 0. 2 Assuming 0 < t ≤ √ π 2, we choose cos(t) = 16x2 −1 . 4 |x | Since csc(t) ≥ 1 in this case and csc(t) = 4x, we have 4x ≥ 1 or x ≥ 1 . Hence, in this case, |x| = x so cos(t) = 4 √ 16x2 −1 4x get x ≤ . For π < t ≤ 1 −4 3π 2, √ we choose cos(t) = − 16x2 −1 4|x| √ < 0 so |x| = −x. This leads to cos(t) = − √ 16x2 −1 4|x| = and since csc(t) ≤ −1 here, we 16x2 −1 4 |x | √ =− 16x2 −1 4(−x) √ = 16x2 −1 in 4 √x 2 −1 = 16x . 4x this case, too. Hence, we have established that, in all cases: cos(arccsc(4x)) Since the domain of arccsc(x) requires |x| ≥ 1, arccsc(4x) requires |4x| ≥ 1 or, using 1 Theorem 2.3, for x to lie in −∞, − 4 ∪ 1 , ∞ . Since we found no additional restrictions 4 √ on t, cos(arccsc(4x)) = 16x2 −1 4x for all x in −∞, − 1 ∪ 4 1 4, ∞ . 714 10.6.3 Foundations of Trigonometry Using a Calculator to Approximate Inverse Function Values. In the sections to...
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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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