Stitz-Zeager_College_Algebra_e-book

# Since 2255 isnt the cosine of one of the common

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Unformatted text preview: ) + (A − C ) cos(2θ)] + B sin(2θ) 2C = [(A + C ) − (A − C ) cos(2θ)] − B sin(2θ) Next, we try to make sense of the product (2A )(2C ) = {[(A + C ) + (A − C ) cos(2θ)] + B sin(2θ)} {[(A + C ) − (A − C ) cos(2θ)] − B sin(2θ)} 5 We hope that someday you get to see why this works the way it does. 11.6 Hooked on Conics Again 833 We break this product into pieces. First, we use the diﬀerence of squares to multiply the ‘ﬁrst’ quantities in each factor to get [(A + C ) + (A − C ) cos(2θ)] [(A + C ) − (A − C ) cos(2θ)] = (A + C )2 − (A − C )2 cos2 (2θ) Next, we add the product of the ‘outer’ and ‘inner’ quantities in each factor to get −B sin(2θ) [(A + C ) + (A − C ) cos(2θ)] +B sin(2θ) [(A + C ) − (A − C ) cos(2θ)] = −2B (A − C ) cos(2θ) sin(2θ) The product of the ‘last’ quantity in each factor is (B sin(2θ))(−B sin(2θ)) = −B 2 sin2 (2θ). Putting all of this together yields 4A C = (A + C )2 − (A − C )2 cos2 (2θ) − 2B (A − C ) cos(2θ) sin(2θ) − B 2 sin2 (2θ) θ) From cot(2θ) = A−C...
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