Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 47


The exact value of the trigonometric function $\sin \frac{\theta }{2}$ is $\frac{\sqrt{10}}{10}$.

Work Step by Step

Recall the figure mentioned in the problem. The figure shows the right-angle triangle. In this triangle, the base is $4$, the perpendicular is $3$, and the hypotenuse is $5$. Calculate the value of $\sin \frac{\theta }{2}$. Recall the half angle formula. $\begin{align} & \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}} \\ & =\sqrt{\frac{1-\left( \frac{\text{base}}{\text{hypotenuse}} \right)}{2}} \end{align}$ Substitute $4$ for the base and $5$ for the hypotenuse. $\begin{align} & \sin \frac{\theta }{2}=\sqrt{\frac{1-\left( \frac{\text{4}}{\text{5}} \right)}{2}} \\ & =\sqrt{\frac{1}{10}} \\ & =\frac{1}{\sqrt{10}}\times \frac{\sqrt{10}}{\sqrt{10}} \\ & =\frac{\sqrt{10}}{10} \end{align}$ Therefore, the exact value of the trigonometric function $\sin \frac{\theta }{2}$ is $\frac{\sqrt{10}}{10}$.
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