Stitz-Zeager_College_Algebra_e-book

Our nal answer is 22 2 2 c x 2 cos x 1 3 3 2 2

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Unformatted text preview: ator of the right hand side is 1 + cos(θ). This suggests the strategy of starting with the left hand side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 − cos(θ) = sin(θ) (1 + cos(θ)) · (1 − cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 − cos(θ))(1 + cos(θ)) = sin(θ)(1 + cos(θ)) 1 − cos2 (θ) = sin(θ)(1 + cos(θ)) sin2 (θ) = sin( $ $$θ )(1 + cos(θ )) sin( $ $$θ ) sin(θ ) = 1 + cos(θ) sin(θ) The reader is encouraged to study the techniques demonstrated in Example 10.3.3. Simply memorizing the fundamental identities is not enough to guarantee success in verifying more complex identities; a fair amount of Algebra is usually required as well. Be on the lookout for opportunities to simplify complex fractions and get common denominators. Another common technique is to exploit so-called ‘Pythagorean Conjugates.’ These are factors such as 1 − sin(θ) and 1 + sin(θ), 10.3 The Six Circular Functions and Fundamental Identities 643...
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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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