## Trigonometry (11th Edition) Clone

The two points of intersection are: $(\frac{4+\sqrt{2}}{2}, \frac{\pi}{4})$ and $(\frac{4-\sqrt{2}}{2}, \frac{5\pi}{4})$
$r = 2+sin~\theta$ $r = 2+cos~\theta$ To find the points of intersection, we can equate the expressions for $r$: $2+sin~\theta = 2+cos~\theta$ $sin~\theta = cos~\theta$ $\frac{sin~\theta}{cos~\theta} = 1$ $tan~\theta = 1$ $\theta = \frac{\pi}{4}, \frac{5\pi}{4}$ We can find $r$ when $\theta = \frac{\pi}{4}$: $r = 2+sin~\theta$ $r = 2+sin~\frac{\pi}{4}$ $r = 2+\frac{\sqrt{2}}{2}$ $r = \frac{4+\sqrt{2}}{2}$ We can find $r$ when $\theta = \frac{5\pi}{4}$: $r = 2+sin~\theta$ $r = 2+sin~\frac{5\pi}{4}$ $r = 2-\frac{\sqrt{2}}{2}$ $r = \frac{4-\sqrt{2}}{2}$ The two points of intersection are: $(\frac{4+\sqrt{2}}{2}, \frac{\pi}{4})$ and $(\frac{4-\sqrt{2}}{2}, \frac{5\pi}{4})$