Answer
The slope of the tangent line at that point is equal to: $$1$$
Work Step by Step
1. Write the formula for the slope of the tangent line: $$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}sin(\theta) + rcos(\theta)}{\frac{dr}{d\theta}cos(\theta) - rsin(\theta)} $$
2. Calculate $\frac{dr}{d\theta}$ and substitute the equation into the formula:
$\frac{dr}{d\theta} = \frac{d(cos(2\theta))}{d\theta} = -2sin(2\theta)$
$\frac{dy}{dx} = \frac{(-2sin(2\theta))sin(\theta) + (cos(2\theta))cos(\theta)}{(-2sin(2\theta))cos(\theta) - (cos(2\theta))sin(\theta)} $
3. Calculate the slope when $\theta = \pi/4$
$\frac{dy}{dx} = \frac{(-2sin(2\frac{\pi}4))sin(\frac{\pi}4) + (cos(2\frac{\pi}4))cos(\frac{\pi}4)}{(-2sin(2\frac{\pi}4))cos(\frac{\pi}4) - (cos(2\frac{\pi}4))sin(\frac{\pi}4)} $
$\frac{dy}{dx} = \frac{(-2(1))(\frac{\sqrt 2} 2) + (0)cos(\frac{\pi}4)}{(-2(1))\frac {\sqrt 2}2 - (0)sin(\frac{\pi}4)} = \frac{- \sqrt 2}{- \sqrt 2} = 1$
