# What is the Cartesian form of (-4,(-21pi)/4))?

Jul 26, 2018

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

#### Explanation:

We have the coordinate $\left(- 4 , \frac{- 21 \pi}{4}\right)$ in polar form.

Coordinates in polar form have the standard form (color(green)(r), color(purple)(Θ)).

To convert from polar form to Cartesian form, we use the following formulas:

• color(red)(x) = color(green)(r)coscolor(purple)(Θ)
• color(blue)(y) = color(green)(r)sincolor(purple)(Θ)

Now, let's plug stuff in. We know $\textcolor{g r e e n}{r} = - 4$ and color(purple)(Θ) = (-21pi)/4

$\textcolor{red}{x} = \left(- 4\right) \cdot \cos \left(\frac{- 21 \pi}{4}\right)$

$\textcolor{red}{x} = \left(- 4\right) \cdot \left(- \frac{\sqrt{2}}{2}\right)$

$\textcolor{red}{x} = 2 \sqrt{2}$

$\textcolor{b l u e}{y} = \left(- 4\right) \cdot \sin \left(\frac{- 21 \pi}{4}\right)$

$\textcolor{b l u e}{y} = \left(- 4\right) \cdot \left(\frac{\sqrt{2}}{2}\right)$

$\textcolor{b l u e}{y} = - 2 \sqrt{2}$

We get $\textcolor{red}{x} = 2 \sqrt{2}$ and $\textcolor{b l u e}{y} = - 2 \sqrt{2}$, making our Cartesian coordinate $\left(2 \sqrt{2} , - 2 \sqrt{2}\right) .$

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

#### Explanation:

Cartesian coordinates $\left(x , y\right)$ of given point $\left(r , \setminus \theta\right) \setminus \equiv \left(- 4 , \frac{- 21 \setminus \pi}{4}\right)$ are given as

$x = r \setminus \cos \setminus \theta$

$= - 4 \setminus \cos \left(- \frac{21 \setminus \pi}{4}\right)$

$= - 4 \setminus \cos \left(\frac{5 \setminus \pi}{4}\right)$

$= - 4 \left(- \frac{1}{\setminus} \sqrt{2}\right)$

$= 2 \setminus \sqrt{2}$

$y = r \setminus \sin \setminus \theta$

$= - 4 \setminus \sin \left(- \frac{21 \setminus \pi}{4}\right)$

$= 4 \setminus \sin \left(\frac{5 \setminus \pi}{4}\right)$

$= 4 \left(- \frac{1}{\setminus} \sqrt{2}\right)$

$= - 2 \setminus \sqrt{2}$

hence the cartesian coordinates of given point are

$\left(x , y\right) \setminus \equiv \left(2 \setminus \sqrt{2} , - 2 \setminus \sqrt{2}\right)$