Stitz-Zeager_College_Algebra_e-book

Hence arccos v w vw 1 v v 1 w v w vw using theorem

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Unformatted text preview: tion 7.5, so we proceed here under the assumption that B = 0. We rotate the xy -axes counter-clockwise through an angle θ which satisfies cot(2θ) = A−C to produce an equation with no x y -term in accordance with B Theorem 11.10: A (x )2 + C (y )2 + Dx + Ey + F = 0. In this form, we can invoke Exercise 10 in Section 7.5 once more using the product A C . Our goal is to find the product A C in terms of the coefficients A, B and C in the original equation. To that end, we make the usual substitutions x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) into Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0. We leave it to the reader to show that, after gathering like terms, the coefficient A on (x )2 and the coefficient C on (y )2 are A = A cos2 (θ) + B cos(θ) sin(θ) + C sin2 (θ) C = A sin2 (θ) − B cos(θ) sin(θ) + C cos2 (θ) In order to make use of the condition cot(2θ) = A−C , we rewrite our formulas for A and C using B the power reduction formulas. After some regrouping, we get 2A = [(A + C...
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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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