Stitz-Zeager_College_Algebra_e-book

Using a cofunction identity along with the even

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Unformatted text preview: n(θ) cos(θ) √ = 1, which gives cos2 (θ) = 4 , or cos(θ) = ± 2 5 5 . Since 5 √ 25 5. = 3. Be careful - this does NOT mean we can take sin(θ) = 3 and sin( cos(θ) = 1. Instead, from cos(θ) = 3 we get: sin(θ) = 3 cos(θ). To relate cos(θ) and sin(θ), we θ) once again employ the Pythagorean Identity, cos2 (θ) + sin2 (θ) = 1. Solving sin(θ) = 3 cos(θ) 1 for cos(θ), we find cos(θ) = 3 sin(θ). Substituting this into the Pythagorean Identity, we find sin2 (θ) + 1 3 sin(θ) 2 = 1. Solving, we get sin2 (θ) = 9 10 √ 10 so sin(θ) = ± 3 10 . Since θ is a √ 10 Quadrant III angle, we know sin(θ) < 0, so our final answer is sin(θ) = − 3 10 . While the Reciprocal and Quotient Identities presented in Theorem 10.6 allow us to always reduce problems involving secant, cosecant, tangent and cotangent to problems involving cosine and sine, it is not always convenient to do so.2 It is worth taking the time to memorize the tangent and cotangent values of the common angles summar...
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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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