Calculus: Early Transcendentals 8th Edition

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

APPENDIX D - Trigonometry - D Exercises - Page A 33: 59

Answer

$sin(x+y)=\frac{1}{15}(4+6\sqrt 2)$

Work Step by Step

Evaluate the expression $sin(x+y)$ Given: $sinx=\frac{1}{3}$ and $secy=\frac{5}{4}$ $sin(x+y)=sinxcosy+cosxsiny$ ...(1) Thus, $sinx=\frac{1}{3}$ gives opp =1 , hyp = 3 and adj $=\sqrt {3^{2}-1^{2}}=2\sqrt 2$ Therefore, $cos x=\frac{2\sqrt 2}{3}$ Now, $secy=\frac{5}{4}$ gives hyp =5 , adj =4 and opp $=\sqrt {5^{2}-3^{2}}=3$ Therefore, $siny=\frac{3}{5}$ and $cosy=\frac{4}{5}$ Equation (1) becomes $sin(x+y)=(\frac{1}{3})(\frac{4}{5})+(\frac{2\sqrt 2}{3})(\frac{3}{5})$ $=\frac{4}{15}+\frac{6\sqrt 2}{15}$ Hence, $sin(x+y)=\frac{1}{15}(4+6\sqrt 2)$
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