Precalculus (6th Edition) Blitzer

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

Chapter 6 - Cumulative Review Exercises - Page 801: 8


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Work Step by Step

As we know that, $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B $. Therefore: $\cos \left( \theta +\frac{3\pi }{2} \right)$, can be written as: $\begin{align} & \cos \left( \theta +\frac{3\pi }{2} \right)=\cos \theta \cos \left( \frac{3\pi }{2} \right)-\sin \theta \sin \left( \frac{3\pi }{2} \right) \\ & =\cos \theta \left( 0 \right)-\sin \theta \left( -1 \right) \\ & =0+\sin \theta \\ & =\sin \theta \end{align}$ Therefore, from (1) we get; $\cos \left( \theta +\frac{3\pi }{2} \right)=\sin \theta $
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