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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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