Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 661: 103

Answer

See the full explanation below.

Work Step by Step

Let us consider the left side of the given expression: $\sin \left( {{30}^{{}^\circ }}+{{60}^{{}^\circ }} \right)$ Now, evaluate the expression and substitute the appropriate values: $\begin{align} & \sin \left( {{30}^{{}^\circ }}+{{60}^{{}^\circ }} \right)=\sin {{90}^{{}^\circ }} \\ & =1 \end{align}$ And consider the right side of the provided expression: $\sin {{30}^{{}^\circ }}+\sin {{60}^{{}^\circ }}$ Now, evaluate the expression and substitute the appropriate values: $\begin{align} & \sin {{30}^{{}^\circ }}+\sin {{60}^{{}^\circ }}=\frac{1}{2}+\frac{\sqrt{3}}{2} \\ & =\frac{1+\sqrt{3}}{2} \end{align}$ Thus, the left side of the given expression is not equal to the right side, which is $\sin \left( {{30}^{{}^\circ }}+{{60}^{{}^\circ }} \right)$ or $\sin {{90}^{{}^\circ }}$ and equal to $\sin {{30}^{{}^\circ }}+\sin {{60}^{{}^\circ }}$. Hence, they are not equal. (b) Let us consider the left side of the given expression: $\sin \left( 30{}^\circ +60{}^\circ \right)$ Now, evaluate the expression and substitute the appropriate values: $\begin{align} & \sin \left( {{30}^{{}^\circ }}+{{60}^{{}^\circ }} \right)=\sin {{90}^{{}^\circ }} \\ & =1 \end{align}$ Now, consider the right side of the provided expression: $\sin 30{}^\circ \cos 60{}^\circ +\cos 30{}^\circ \sin 60{}^\circ $ And, evaluate the expression and substitute the appropriate values: $\begin{align} & \sin 30{}^\circ \cos 60{}^\circ +\cos 30{}^\circ \sin 60{}^\circ ~=\left( \frac{1}{2} \right)\left( \frac{1}{2} \right)+\left( \frac{\sqrt{3}}{2} \right)\left( \frac{\sqrt{3}}{2} \right) \\ & =\frac{1}{4}+\frac{3}{4} \\ & =1 \end{align}$ Thus, the left side of the given expression is equal to the right side, which is $\sin \left( 30{}^\circ +60{}^\circ \right)$ or $\sin 90{}^\circ $ and equal to $\sin 30{}^\circ \cos 60{}^\circ +\cos 30{}^\circ \sin 60{}^\circ $. Hence, they are equal.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.