Stitz-Zeager_College_Algebra_e-book

We are after 2 cot arccsc 3 cott so we use the

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Unformatted text preview: Theorem 10.18 1 − cos(2θ) 2 sin2 (θ) cos2 (θ) = = = 1 + cos(2θ) 2 1 1 − cos2 (2θ) 4 11 − cos2 (2θ) 44 Next, we apply the power reduction formula to cos2 (2θ) to finish the reduction sin2 (θ) cos2 (θ) = = = = 1 4 1 4 1 4 1 8 1 cos2 (2θ) 4 1 1 − cos(2(2θ)) − 4 2 11 − + cos(4θ) 88 1 + cos(4θ) 8 − Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we θ apply the Power Reduction Formula to cos2 2 664 Foundations of Trigonometry cos2 θ 2 = θ 2 1 + cos 2 2 1 + cos(θ) . 2 = θ We can obtain a formula for cos 2 by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below. Theorem 10.19. Half Angle Formulas: For all applicable angles θ, • cos θ 2 =± 1 + cos(θ) 2 • sin θ 2 =± 1 − cos(θ) 2 • tan θ 2 =± 1 − cos(θ) 1 + cos(θ) where the choice of ± depends on the quadrant in which the terminal side of θ lies. 2 Example 10.4.5. 1. U...
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