## Trigonometry 7th Edition

$2\sqrt2$
The terminal side of the angle $\dfrac{-\pi}{4}$ is in Quadrant IV. This means its cosine value is positive. 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 an $\frac{\pi}{4}$ 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 cosine values, except possibly for their signs. $\dfrac{\pi}{4}$ is a special angle whose cosine value if $\dfrac{\sqrt2}{2}$. Since $-\dfrac{\pi}{4}$ in Quadrant IV, its cosine is positive. Therefore $\cos{(-\dfrac{\pi}{4})} = \dfrac{\sqrt2}{2}$. Substitute this value into the given expression to obtain: $4 \cos{\left(-\dfrac{\pi}{4}\right)}=4 \cdot \dfrac{\sqrt2}{2} = 2\sqrt2$