Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX D - Trigonometry - D Exercises: 86

Answer

$ cos (\alpha+\beta)= cos \alpha cos\beta- sin \alpha sin\beta$

Work Step by Step

Apply the distance formula for $(cos \alpha, sin\alpha),(cos \beta, sin\beta)$ $c^{2}=(cos \beta-cos \alpha)^{2}+(sin \beta-sin \alpha)^{2}$ $c^{2}=(cos \beta-cos \alpha)^{2}+(sin \beta-sin \alpha)^{2}$ $c^{2}=2-2 cos \alpha cos\beta-2 sin \alpha sin\beta$ Using the law of cosines, we have $c^{2}=a^{2}+b^{2}-2ab cos \theta$ Here, $c=c, a=1, b=1, \theta=\alpha-\beta$ Thus, $c^{2}=2-2 cos (\alpha-\beta)$ ...(2) From equations (1) and (2), we have $2-2 cos (\alpha-\beta)=2-2 cos \alpha cos\beta-2 sin \alpha sin\beta$ $2 cos (\alpha-\beta)=2 cos \alpha cos\beta+2 sin \alpha sin\beta$ $ cos (\alpha-\beta)= cos \alpha cos\beta+ sin \alpha sin\beta$ Change the sign of $\beta$ with $-\beta$. $ cos (\alpha+\beta)= cos \alpha cos(-\beta)+ sin \alpha sin(-\beta)$ Hence, $ cos (\alpha+\beta)= cos \alpha cos\beta- sin \alpha sin\beta$
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