# Chapter 3 - Section 3.2 - Radians and Degrees - 3.2 Problem Set - Page 133: 68

$-2\sqrt2$

#### Work Step by Step

The terminal side of the angle $\dfrac{-\pi}{4}$ is in Quadrant IV. The reference angle of an angle is equal to the smallest acute angle that its terminal side makes with the x-axis. The terminal side of $-\frac{\pi}{4}$ makes a $\frac{\pi}{4}$ angle with the positive x-axis. Thus, the reference angle of the given angle is $\dfrac{\pi}{4}$. Recall that an angle and its reference angle have the same sine values, except possibly for their signs. $\dfrac{\pi}{4}$ is a special angle whose sine value if $\dfrac{\sqrt2}{2}$. Since $-\dfrac{\pi}{4}$ in Quadrant IV, its sine is negative. Therefore $\sin{(-\dfrac{\pi}{4})} = -\dfrac{\sqrt2}{2}$. Substitute this value into the given expression to obtain: $4 \sin{\left(-\dfrac{\pi}{4}\right)}=4 \cdot \dfrac{-\sqrt2}{2} = -2\sqrt2$

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