Precalculus (6th Edition) Blitzer

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

Chapter 10 - Section 10.5 - The Binomial Theorum - Exercise Set - Page 1094: 89

Answer

The value is $\frac{\sqrt{3}-1}{2\sqrt{2}}$.

Work Step by Step

Use the equation: $\cos 75{}^\circ =\cos \left( 120{}^\circ -45{}^\circ \right)$ Applying the formula $\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b $ We get: $\begin{align} & \cos \left( 120{}^\circ -45{}^\circ \right)=\cos 120{}^\circ \cos 45{}^\circ +\sin 120{}^\circ \sin 45{}^\circ \\ & =-\frac{1}{2}\times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} \\ & =\frac{\sqrt{3}-1}{2\sqrt{2}} \end{align}$ Hence, $\cos 75{}^\circ =\frac{\sqrt{3}-1}{2\sqrt{2}}$.
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