Stitz-Zeager_College_Algebra_e-book

3 example 1193 let l1 be the line y m1 x b1 and let l2

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Unformatted text preview: , we get cos(2θ) = A−C , or (A − C ) sin(2θ) = B cos(2θ). We use this substitution B B sin(2 twice along with the Pythagorean Identity cos2 (2θ) = 1 − sin2 (2θ) to get 4A C = (A + C )2 − (A − C )2 cos2 (2θ) − 2B (A − C ) cos(2θ) sin(2θ) − B 2 sin2 (2θ) = (A + C )2 − (A − C )2 1 − sin2 (2θ) − 2B cos(2θ)B cos(2θ) − B 2 sin2 (2θ) = (A + C )2 − (A − C )2 + (A − C )2 sin2 (2θ) − 2B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C )2 − (A − C )2 + [(A − C ) sin(2θ)]2 − 2B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C )2 − (A − C )2 + [B cos(2θ)]2 − 2B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C )2 − (A − C )2 + B 2 cos2 (2θ) − 2B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C )2 − (A − C )2 − B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C )2 − (A − C )2 − B 2 cos2 (2θ) + sin2 (2θ) = (A + C )2 − (A − C )2 − B 2 = A2 + 2AC + C 2 − A2 − 2AC + C 2 − B 2 = 4AC − B 2 Hence, B 2 − 4AC = −4A C , so the quantity B 2 − 4AC h...
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