# What is the Cartesian form of (-16,(-23pi)/4))?

Nov 15, 2016

(-$\frac{16}{\sqrt{2}} , - \frac{16}{\sqrt{2}}$)

#### Explanation:

The given point is r= -16 and $\theta = \frac{- 23 \pi}{4}$

The cartesian coordinates would be x= $r \cos \theta$ and y = $r \sin \theta$

Accordingly, x= $- 16 \cos \left(\frac{- 23 \pi}{4}\right)$ and y= $- 16 \sin \left(\frac{- 23 \pi}{4}\right)$

Hence x= -16 cos $- \left(6 \pi - \frac{\pi}{4}\right)$ and y=-16sin $- \left(6 \pi - \frac{\pi}{4}\right)$

x= -16cos $\left(6 \pi - \frac{\pi}{4}\right)$ and y= 16sin $\left(6 \pi - \frac{\pi}{4}\right)$

x=-16 cos$\left(- \frac{\pi}{4}\right)$ and y= 16 sin $- \left(\frac{\pi}{4}\right)$

x= -16 cos $\frac{\pi}{4}$ and y= -16 $\sin \frac{\pi}{4}$

x=-$\frac{16}{\sqrt{2}}$ and y= -$\frac{16}{\sqrt{2}}$