Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Mid-Chapter Check Point - Page 684: 23


The required value is $-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right)$

Work Step by Step

By using the sum formula $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$ and trigonometric values to compute the value of the given expression, we get: $\begin{align} & \sin \left( \frac{3\pi }{4}+\frac{5\pi }{6} \right)=\sin \frac{3\pi }{4}\cos \frac{5\pi }{6}+\cos \frac{3\pi }{4}\sin \frac{5\pi }{6} \\ & =\frac{\sqrt{2}}{2}.\frac{-\sqrt{3}}{2}+\frac{-\sqrt{2}}{2}.\frac{1}{2} \\ & =-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ & =-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right) \end{align}$ Hence, the required value is $-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right)$
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