Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 55



Work Step by Step

Use a graphing utility to graph the given expression. (Refer to the graph below,) Notice that the graph is identical with the graph of $\tan{2x}$. This means that $\dfrac{2\tan{x}}{2-\sec^2{x}}=\tan{(2x)}$. Work on the left side of the equation above by writing each function using $\sin{x}$ and $\cos{x}$ only: \begin{align*} \require{cancel} \dfrac{2\tan{x}}{2-\sec^2{x}}&=\dfrac{2\left(\frac{\sin{x}}{\cos{x}}\right)}{2-\frac{1}{\cos^2{x}}}\\\\ &=\dfrac{\frac{2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}}{\cos^2{x}}-\frac{1}{\cos^2{x}}}\\\\ &=\dfrac{\frac{2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}-1}{\cos^2{x}}}\\\\ &=\frac{2\sin{x}}{\cos{x}} \cdot \frac{\cos^2{x}}{\cos^2{x}-1}\\\\ &=\frac{2\sin{x}}{\cancel{\cos{x}}} \cdot \frac{\cos^\cancel{2}{x}}{\cos^2{x}-1}\\\\ &=2\sin{x} \cdot \frac{\cos{x}}{\cos^2{x}-1}\\\\ &=\frac{2\sin{x}\cos{x}}{\cos^2{x}-1}\\\\ \end{align*} Recall: (1) $2\cos^2{A}-1=\cos{(2A)}\\$ (2) $2\sin{A}\cos{A}=\sin{(2A)}$ Use the rules above to obtain: \begin{align*} \require{cancel} \dfrac{2\tan{x}}{2-\sec^2{x}}&=\frac{2\sin{x}\cos{x}}{\cos^2{x}-1}\\\\ &=\dfrac{\sin{2x}}{\cos{2x}}\\\\ &=\tan{2x} \end{align*}
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