Stitz-Zeager_College_Algebra_e-book

Find a sinsuoid which models the height h of the

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Unformatted text preview: gardless. Second, as we have mentioned, there is no universally accepted range of the arcsecant function. For that reason, we adopt the advice given in Section 10.3 and convert 3 this to the cosine problem cos(x) = − 5 . Adopting an angle approach, we consider the equation cos(θ) = − 3 and note our solutions lie in Quadrants II and III. The reference 5 angle α satisfies 0 < α < π with cos(α) = 3 . We look to the arccosine function for help. 2 5 3 3 The real number t = arccos 5 satisfies 0 < t < π and cos(t) = 5 , so our reference angle 2 is α = arccos 3 radians. Proceeding as before, we find the Quadrant II solutions to be 5 θ = π − α + 2πk = π − arccos 3 + 2πk for integers k . In Quadrant III, one angle with 5 3 reference angle α is π + α, so our solutions here are θ = π + α + 2πk = π + arccos 5 + 2πk for 12 When in doubt, write them out! 10.6 The Inverse Trigonometric Functions 719 integers k . Passing back to real numbers, we state our solutions as t = π − arccos or t = π + arccos 3 + 2πk for integers k . 5 y + 2πk y 1 3 5 1 α 1 x α 1 x It is natural to wonder if it is possible to...
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