Calculus: Early Transcendentals 8th Edition

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

Diagnostic Tests - D Diagnostic Tests: Trigonometry - Page xxx: 6

Answer

$\frac{1}{15}(4+6\sqrt 2)$

Work Step by Step

sin x = $\frac{1}{3}$ Pythagoras theorem: $1^{2}+{a}^{2}=3^{2}$ a=$\sqrt 8$ cos x = $\frac{\sqrt 8}{3}$ sec y = $\frac{5}{4}$ cos y = $\frac{1}{sec y}$=$\frac{4}{5}$ Pythagoras theorem: $b^{2}+{4}^{2}=5^{2}$ b=3 sin y = $\frac{3}{5}$ Expand: sin ( x + y ) = sin x cos y + sin y cos x = $\frac{1}{3}\times\frac{4}{5} + \frac{3}{5}\times\frac{\sqrt 8}{3}$ =$\frac{4+6\sqrt 2}{15}$ =$\frac{1}{15}(4+6\sqrt 2)$
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