Calculus: Early Transcendentals 8th Edition

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

APPENDIX D - Trigonometry - D Exercises: 87

Answer

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

Work Step by Step

Use the addition formula for cosine and sine identities. $ cos (\alpha+\beta)= cos \alpha cos\beta- sin \alpha sin\beta$ Change $\alpha$ with $(\pi/2-\alpha)$ $ cos [(\pi/2-\alpha)+\beta]= cos (\pi/2-\alpha) cos\beta- sin (\pi/2-\alpha) sin\beta$ $ cos [\pi/2-(\alpha-\beta)]= cos (\pi/2-\alpha) cos\beta- sin (\pi/2-\alpha) sin\beta$ Hence, $ sin (\alpha-\beta)=sin \alpha cos\beta- cos\alpha sin\beta$
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