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: 4

Answer

The exact value of the trigonometric function $\sin 2\alpha $ is $\frac{336}{625}$.

Work Step by Step

Calculate the value of the hypotenuse. For the right angle triangle. $\text{hypotenuse}=\sqrt{\text{perpendicula}{{\text{r}}^{2}}+\text{bas}{{\text{e}}^{2}}}$ Substitute $24$ for the base and $7$ for the perpendicular. $\begin{align} & \text{hypotenuse}=\sqrt{{{\text{7}}^{2}}+\text{2}{{\text{4}}^{2}}} \\ & =\sqrt{625} \\ & =25 \end{align}$ Calculate the value of $\sin 2\alpha $. Recall the double angle formula. $\begin{align} & \sin 2\alpha =2\sin \alpha \cos \alpha \\ & =2\left( \frac{\text{perpendicular}}{\text{hypotenuse}} \right)\left( \frac{\text{base}}{\text{hypotenuse}} \right) \end{align}$ Substitute $24$ for the base, $7$ for the perpendicular and $25$ for the hypotenuse. $\begin{align} & \sin 2\alpha =2\left( \frac{\text{7}}{\text{25}} \right)\left( \frac{\text{24}}{\text{25}} \right) \\ & =\frac{336}{625} \end{align}$ Therefore, the exact value of the trigonometric function $\sin 2\alpha $ is $\frac{336}{625}$.
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