Stitz-Zeager_College_Algebra_e-book

Common sense suggests that if we release the object

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Unformatted text preview: tly apply Theorem 10.27 to help us find arccot(−2). Let t = arccot(−2). Then t is a real number between 0 and π with cot(t) = −2. Let θ = t radians. Then θ is an angle between 0 and π with cot(θ) = −2. Since cot(θ) < 0, we know θ must be a Quadrant II angle. Consider the reference angle for θ, α, as pictured below. By definition, 0 < α < π and by the Reference Angle Theorem, Theorem 10.2, it follows that 2 cot(α) = 2. By definition, then, α = arccot(2) radians which we can rewrite using Theorem 10.27 as arctan 1 . Since θ + α = π , we have θ = π − α = π − arctan 1 ≈ 2.6779 radians. 2 2 Since θ = t radians, we have arccot(−2) ≈ 2.6779. 10.6 The Inverse Trigonometric Functions 715 y 1 θ α 1 x 4. If the range of arccsc(x) is taken to be − π , 0 ∪ 0, π , we can use Theorem 10.28 to get 2 2 arccsc(−5) = arcsin − 1 ≈ −0.2014. If, on the other hand, the range of arccsc(x) is taken 5 π to be 0, π ∪ π , 32 , then we proceed as in the p...
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