Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.1 - Using Fundamental Identities - 7.1 Exercises - Page 514: 53


$\sqrt {9-x^2}=3~sin~θ$

Work Step by Step

$sin^2θ+cos^2θ=1$ $cos^2θ=1−sin^2θ$ $\sqrt {9-x^2}=\sqrt {9-(3~cosθ)^2}=\sqrt {9-9~cos^2θ}=\sqrt {9(1-cos^2θ)}=3\sqrt {sin^2θ}=±3~sin~θ$ But, since $0\ltθ\lt\frac{\pi}{2}$ (Quadrant I): $\sqrt {9-x^2}=3~sin~θ$
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