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: 15

Answer

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

Work Step by Step

Rewrite the expression for sine of $105{}^\circ $ as the sum of two angles as, $\sin 105{}^\circ =\sin \left( 60{}^\circ +45{}^\circ \right)$ Use the sum formula of sines and evaluate the modified expression as, $\sin \left( 60{}^\circ +45{}^\circ \right)=\sin 60{}^\circ \cos 45{}^\circ +\cos 60{}^\circ \sin 45{}^\circ $ Substitute the values $\cos 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\cos 60{}^\circ =\frac{1}{2},\text{ }\sin 45{}^\circ =\frac{1}{\sqrt{2}},\text{ and }\sin 30{}^\circ =\frac{\sqrt{3}}{2}$. $\begin{align} & \sin \left( 60{}^\circ +45{}^\circ \right)=\left( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} \right)+\left( \frac{1}{2}\times \frac{1}{\sqrt{2}} \right) \\ & =\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \\ & =\frac{\sqrt{3}+1}{2\sqrt{2}}\\ &=\frac{\sqrt 6+\sqrt 2}{4} \end{align}$ Hence, the exact value of $\sin 105{}^\circ $ is equivalent to $\frac{\sqrt 6+\sqrt 2}{4}$.
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