Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 14


The exact value of $\sin \left( 60{}^\circ -45{}^\circ \right)$ is $\frac{\sqrt{3}-1}{2\sqrt{2}}$.

Work Step by Step

Use the difference formula of sines and evaluate the expression as, $\sin \left( 60{}^\circ -45{}^\circ \right)=\sin 60{}^\circ \cos 45{}^\circ -\cos 60{}^\circ \sin 45{}^\circ $ Substitute the values $\cos 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\cos 60{}^\circ =\frac{1}{2},\text{ }\sin 45{}^\circ =\frac{1}{\sqrt{2}},\text{ and }\sin 30{}^\circ =\frac{\sqrt{3}}{2}$. $\begin{align} & \sin \left( 60{}^\circ -45{}^\circ \right)=\left( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2}\times \frac{1}{\sqrt{2}} \right) \\ & =\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}} \\ & =\frac{\sqrt{3}-1}{2\sqrt{2}} \end{align}$ Hence, the exact value of $\sin \left( 60{}^\circ -45{}^\circ \right)$ is equivalent to $\frac{\sqrt{3}-1}{2\sqrt{2}}$.
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