Stitz-Zeager_College_Algebra_e-book

Note that whether we graph y g x using the quarter

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Unformatted text preview: (1 − sin(θ))(1 + sin(θ)) (1 + sin(θ))(1 − sin(θ)) = 3 + 3 sin(θ) 3 − 3 sin(θ) − 1 − sin2 (θ) 1 − sin2 (θ) = (3 + 3 sin(θ)) − (3 − 3 sin(θ)) 1 − sin2 (θ) = 6 sin(θ) 1 − sin2 (θ) At this point, it is worth pausing to remind ourselves of our goal. We wish to transform this expression into 6 sec(θ) tan(θ). Using a reciprocal and quotient identity, we find 4 Or, to put to another way, earn more partial credit if this were an exam question! 642 Foundations of Trigonometry 6 sec(θ) tan(θ) = 6 1 cos(θ) sin(θ) cos(θ) . In other words, we need to get cosines in our denomi- nator. To that end, we recall the Pythagorean Identity cos2 (θ) + sin2 (θ) = 1 which we can rewrite as cos2 (θ) = 1 − sin2 (θ). Putting all of this together we finish our proof: 3 3 − 1 − sin(θ) 1 + sin(θ) = 6 sin(θ) 1 − sin2 (θ) = 6 sin(θ) cos2 (θ) =6 1 cos(θ) sin(θ) cos(θ) = 6 sec(θ) tan(θ) 6. It is debatable which side of the identity is more complicated. One thing which stands out is the denominator on the left hand side is 1 − cos(θ), while the numer...
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