# What is the Cartesian form of (-8,(-21pi)/8))?

##### 1 Answer
Feb 21, 2018

$\left(- 4 \sqrt{2 + \sqrt{2}} , - 4 \sqrt{2 - \sqrt{2}}\right)$

#### Explanation:

In order to transform the Polar coordinates of the point $P$$\left(r , \theta\right)$ into the Cartesian coordinates $P$$\left(x , y\right)$, you must look at the right angled triangle made by the origin, $P$ and $P '$, where $P '$ has coordinates $\left(x , 0\right)$ or $\left(r , 0\right)$.

In this triangle,

$\sin \theta = \frac{y}{r}$
and
$\cos \theta = \frac{x}{r}$,

therefore
$x = r \cdot \cos \theta$
and
$y = r \cdot \sin \theta$.

Your example has $\theta = \frac{- 21 \pi}{8}$ and $r = - 8$, so

$x = - 8 \cdot \sin \left(\frac{- 21 \pi}{8}\right)$
$y = - 8 \cdot \cos \left(\frac{- 21 \pi}{8}\right)$

Using the property that the $\sin$ and $\cos$ functions are periodic with period $2 n \pi$, where n is an integer.

This means that for any $a$,

$\sin a = \sin \left(a + 2 n \pi\right)$
$\cos a = \cos \left(a + 2 n \pi\right)$.

$\sin \left(\frac{- 21 \pi}{8}\right) = \sin \left(\frac{- 16 \pi}{8} + \frac{- 5 \pi}{8}\right) = \sin \left(- 2 \pi + \frac{- 5 \pi}{8}\right)$

So $\sin \left(\frac{- 21 \pi}{8}\right) = \sin \left(\frac{- 5 \pi}{8}\right)$.

$\sin \left(\frac{- 5 \pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$
and
$\cos \left(\frac{- 5 \pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$.

Back to the system of equations :

$x = - 4 \sqrt{2 + \sqrt{2}}$

$y = - 4 \sqrt{2 - \sqrt{2}}$.