Stitz-Zeager_College_Algebra_e-book

If the object is released above or below the

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Unformatted text preview: c(x) → π + ; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if 0 < t ≤ – arccsc(x) = arcsin 1 x π 2 or π < t ≤ 3π 2 and csc(t) = x for x ≥ 1 onlyc – csc (arccsc(x)) = x provided |x| ≥ 1 – arccsc(csc(x)) = x provided 0 < x ≤ π 2 or π < x ≤ 3π 2 a . . . assuming the “Calculus Friendly” ranges are used. Compare this with the similar result in Theorem 10.28. c Compare this with the similar result in Theorem 10.28. b Our next example is a duplicate of Example 10.6.3. The interested reader is invited to compare and contrast the solution to each. Example 10.6.4. 1. Find the exact values of the following. 5π 4 (a) arcsec(2) (c) arcsec sec (b) arccsc(−2) (d) cot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 10.6 The Inverse Trigonometric Functions 713 Solution. 1. (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos 1 2 = π. 3 (b) Unfortunately, −2 is not greater to or equal to 1, so we cannot apply Theorem 10.29 to arccsc(−2) and convert this...
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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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