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: 12

Answer

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

Work Step by Step

$$\eqalign{ & \cos \left( { - {{15}^ \circ }} \right) \cr & {\text{Write 1}}{{\text{5}}^ \circ }{\text{ as }}{45^ \circ } - {60^ \circ } \cr & = \cos \left( {{{45}^ \circ } - {{60}^ \circ }} \right) \cr & {\text{Use cosine difference identity}} \cr & = \cos \left( {{{45}^ \circ }} \right)\cos \left( {{{60}^ \circ }} \right) + \sin \left( {{{45}^ \circ }} \right)\sin \left( {{{60}^ \circ }} \right) \cr & {\text{Substitute known values}} \cr & = \left( {\frac{{\sqrt 2 }}{2}} \right)\left( {\frac{1}{2}} \right) + \left( {\frac{{\sqrt 2 }}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) \cr & {\text{Simplifying}} \cr & = \frac{{\sqrt 2 }}{4} + \frac{{\sqrt 6 }}{4} \cr & = \frac{{\sqrt 2 + \sqrt 6 }}{4} \cr} $$
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