# How do you convert the polar coordinate (-2,5pi/4) into cartesian coordinates?

Oct 21, 2015

$\left(- 2 , \frac{5 \pi}{4}\right)$ [polar] $= \left(\sqrt{2} , \sqrt{2}\right)$ [Cartesian]

#### Explanation:

In Polar form:
$\left(- 2 , \frac{5 \pi}{4}\right) = \left(2 , \frac{\pi}{4}\right)$
(draw a diagram if this isn't obvious)

For an angle of $\frac{\pi}{4}$ the "opposite" and "adjacent" sides (i.e. $x$ and $y$) are equal
and since $\sqrt{{x}^{2} + {y}^{2}} =$ the radius $= 2$

$\rightarrow x = y = \sqrt{2}$

Oct 21, 2015

The solution is (-sqrt(2);-sqrt(2)). See explanation for details.

#### Explanation:

The given coordinates are not correct, because $r$ cannot be negative (it is a distance between 2 points and it is either zero or a positive real number).

If the given point was (2;(5pi)/4) then corresponding Carthesian coordinates would be:

$x = r \cos \varphi = 2 \cdot \cos \left(\frac{5 \pi}{4}\right) = 2 \cdot \left(- \cos \left(\frac{\pi}{4}\right)\right)$

$= 2 \cdot \left(- \frac{\sqrt{2}}{2}\right) = - \sqrt{2}$

$y = r \sin \varphi = 2 \cdot \sin \left(\frac{5 \pi}{4}\right) = 2 \cdot \left(- \sin \left(\frac{\pi}{4}\right)\right)$

$= 2 \cdot \left(- \frac{\sqrt{2}}{2}\right) = - \sqrt{2}$

So the Carthesian coordinates are (-sqrt(2);-sqrt(2))