Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.3 Sum and Difference Identities - 7.3 Exercises - Page 679: 56

Answer

$$\frac{{\sqrt 6 - \sqrt 2 }}{4}$$

Work Step by Step

$$\eqalign{ & \sin \left( { - \frac{{13\pi }}{{12}}} \right) \cr & = - \sin \left( {\frac{{13\pi }}{{12}}} \right) \cr & \frac{{13\pi }}{{12}} = \pi + \frac{\pi }{{12}} \cr & = - \sin \left( {\pi + \frac{\pi }{{12}}} \right) \cr & {\text{Recall that sin}}\left( {\pi + \theta } \right) = - \sin \theta ,{\text{ then}} \cr & \sin \frac{{13\pi }}{{12}} = \sin \left( {\frac{\pi }{{12}}} \right) \cr & {\text{Write }}\frac{\pi }{{12}}{\text{ as }}\frac{\pi }{4} - \frac{\pi }{6} \cr & - \sin \frac{{13\pi }}{{12}} = \sin \left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) \cr & {\text{Use sine difference identity}} \cr & - \sin \frac{{13\pi }}{{12}} = \sin \left( {\frac{\pi }{4}} \right)\cos \left( {\frac{\pi }{6}} \right) - \cos \left( {\frac{\pi }{4}} \right)\sin \left( {\frac{\pi }{6}} \right) \cr & {\text{Substitute known values}} \cr & - \sin \frac{{13\pi }}{{12}} = \left( {\frac{{\sqrt 2 }}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) - \left( {\frac{{\sqrt 2 }}{2}} \right)\left( {\frac{1}{2}} \right) \cr & - \sin \frac{{13\pi }}{{12}} = \frac{{\sqrt 6 }}{4} - \frac{{\sqrt 2 }}{4} \cr & - \sin \frac{{13\pi }}{{12}} = \frac{{\sqrt 6 - \sqrt 2 }}{4} \cr} $$
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