Answer
The slope of the tangent line is $~1$
Work Step by Step
$r = cos~2\theta$
$x = r~cos~\theta = cos~2\theta~cos~\theta$
$y = r~sin~\theta = cos~2\theta~sin~\theta$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-2~sin~2\theta~sin~\theta+cos~2\theta~cos~\theta}{-2~sin~2\theta~cos~\theta-cos~2\theta~sin~\theta}$
When $\theta = \frac{\pi}{4}$:
$\frac{dy}{dx} = \frac{-2~sin~\frac{\pi}{2}~sin~\frac{\pi}{4}+cos~\frac{\pi}{2}~cos~\frac{\pi}{4}}{-2~sin~\frac{\pi}{2}~cos~\frac{\pi}{4}-cos~\frac{\pi}{2}~sin~\frac{\pi}{4}}$
$\frac{dy}{dx} = \frac{-2(1)~(\frac{\sqrt{2}}{2})+(0)~(\frac{\sqrt{2}}{2})}{-2~(1)~(\frac{\sqrt{2}}{2})-(0)~(\frac{\sqrt{2}}{2})}$
$\frac{dy}{dx} = \frac{-\sqrt{2}}{-\sqrt{2}}$
$\frac{dy}{dx} = 1$
The slope of the tangent line is $~1$
