# What is the Cartesian form of ( -8 , ( - 15pi)/4 ) ?

Dec 3, 2017

Use the polar conversion formulae $x = r \cos \theta , y = r \sin \theta$. $\left(- 4 \sqrt{2} , - 4 \sqrt{2}\right)$

#### Explanation:

In order to perform the conversion from polar to cartesian coordinates, it helps us to remind ourselves of what the various coordinates represent.

On the xy-plane, the x coordinate simply denotes the x value (i.e. its position along the x-axis), and the y coordinate denotes the y-value (its position along the y-axis)

The polar coordinates of a given point on the xy-plane, however, are represented by $\left(r , \theta\right)$, with $r$ denoting the distance from the origin, and $\theta$ the angle formed between the x-axis and the line connecting the given point to the origin.

For a graphical representation, look here:

https://www.mathsisfun.com/polar-cartesian-coordinates.html

Normally we prefer to deal with a positive r-value.
The length of the segment, we know from the distance formula, will be $\sqrt{{x}^{2} + {y}^{2}}$. We can then make a right triangle with $r$ (the segment) as our hypotenuse, the $x$ coordinate as the length of one side, and the $y$ coordinate as the length of the other side.

If the angle formed between the x-axis and r is represented by $\theta$, then we know that

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

Multiplying both sides by r we get

$y = r \sin \theta , x = r \cos \theta$

Recall several important properties of trigonometric functions to make our conversion process easier:

$\sin \left(- \theta\right) = - \sin \left(\theta\right) , \cos \left(- \theta\right) = \cos \left(\theta\right) , \sin \left(\theta + 2 n \pi\right) = \sin \left(\theta\right) , \cos \left(\theta + 2 n \pi\right) = \cos \left(\theta\right) , \sin \left(\frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
Using this, we can perform our conversion:

$x = - 8 \cos \left(- 15 \frac{\pi}{4}\right) = - 8 \cos \left(- 7 \frac{\pi}{4}\right) = - 8 \cos \left(7 \frac{\pi}{4}\right) = - 8 \frac{\sqrt{2}}{2} = - \frac{8}{\sqrt{2}} = - 4 \sqrt{2}$

$y = - 8 \sin \left(- 15 \frac{\pi}{4}\right) = - 8 \sin \left(- 7 \frac{\pi}{4}\right) = 8 \sin \left(7 \frac{\pi}{4}\right) = 8 \cdot \left(- \frac{1}{\sqrt{2}}\right) = - 4 \sqrt{2}$

The cartesian coordinate form of our point is $\left(x , y\right) = \left(- 4 \sqrt{2} , - 4 \sqrt{2}\right)$

To confirm, we will convert these back into polar coordinates.

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{32 + 32} = \sqrt{64} = \pm 8$
$\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right) = {\tan}^{-} 1 \left(1\right) = \frac{\pi}{4} + n \pi$, which gives us $- 7 \frac{\pi}{4}$ when $n = - 2$, and $- 15 \frac{\pi}{4}$ when $n = - 4$,