{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stitz-Zeager_College_Algebra_e-book

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}