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: 51


See the full explanation below.

Work Step by Step

Let us consider the left side of the given expression, $\sin \,2\alpha $ By using the identity of trigonometry, $\sin \,\left( \alpha +\beta \right)=\sin \,\alpha \,\cos \beta +\cos \,\alpha \,\sin \,\beta $ Now, the above expression can be further simplified as: $\begin{align} & \sin \,2\alpha =\sin \,\left( \alpha +\alpha \right) \\ & =\sin \,\alpha \,\cos \,\alpha +\cos \,\alpha \,\sin \,\alpha \\ & =2\sin \,\alpha \,\cos \,\alpha \end{align}$ Thus, the left side of the provided expression is equal to the right side, which is, $\sin \,2\alpha =2\sin \,\alpha \cos \,\alpha $. Hence, it is proved that $\sin \,2\alpha =2\sin \,\alpha \cos \,\alpha $.
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.