Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 84

Answer

The exact value of the expression is $1$.

Work Step by Step

The $\tan \left( -\frac{11\pi }{4} \right)$ lies in quadrant III. Add $4\pi $ to $-\frac{11\pi }{4}$, to find a positive co-terminal angle less than $2\pi $. $\begin{align} & \alpha =\left( -\frac{11\pi }{4} \right)+4\pi \\ & =\frac{-11\pi +16\pi }{4} \\ & =\frac{5\pi }{4} \end{align}$ The positive acute angle formed by the terminal side of $\theta $ and the x-axis is the reference angle ${\theta }'$. Since, $\frac{5\pi }{4}$ angle lies in quadrant III, subtract $\pi $ from $\frac{5\pi }{4}$ to find the reference angle. $\begin{align} & \theta '=\frac{5\pi }{4}-\pi \\ & =\frac{5\pi -\pi }{4} \\ & =\frac{\pi }{4} \end{align}$ The reference angle of $-\frac{11\pi }{4}$ is $\frac{\pi }{4}$. The function value of the reference angle is $\tan \frac{\pi }{4}=1$ The angle $-\frac{11\pi }{4}$ is in quadrant III and the tangent function is positive in quadrant III. Therefore, $\tan \left( -\frac{11\pi }{4} \right)=\tan \frac{\pi }{4}$ Substitute $1$ for $\tan \frac{\pi }{4}$. $\tan \left( -\frac{11\pi }{4} \right)=1$ The exact value of the expression is $1$.
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