Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 31: 60


$$c^{2} =a^{2}-2 a b \cos \theta+b^{2}$$

Work Step by Step

From the given figure, we apply the Pythagorean Theorem: \begin{aligned} \text {Hypotenuse}^{2}&=\operatorname{side}^{2}+\text { side }^{2}\\ c^{2}&=(a-b \cos \theta)^{2}+(b \sin \theta)^{2}\\ c^{2}&=a^{2}-2(a)(b \cos \theta)+b^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta\\ c^{2}&=a^{2}-2 a b \cos \theta+b^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta\\ c^{2}&=a^{2}-2 a b \cos \theta+b^{2} \end{aligned}
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