# What is the Cartesian form of (-64,(5pi)/12)?

Aug 22, 2016

I got $\left(- 16 \sqrt{6} + 16 \sqrt{2} \text{,} - 16 \sqrt{6} - 16 \sqrt{2}\right)$, and Wolfram Alpha gives the same answer as well:

Since you have two coordinates, $\left(r , \theta\right)$, you are in polar coordinates, a radial direction and a single angle. You can tell because the second coordinate is an angle in radians.

(If you had three coordinates, you would have to specify spherical or cylindrical coordinates.)

In polar coordinates, recall that

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

Therefore, you input the values of $r$ and $\theta$ such that you get your new $\left(x , y\right)$ coordinates.

$\textcolor{b l u e}{\text{(} x}$$\textcolor{b l u e}{,}$ $\textcolor{b l u e}{y \text{)}}$

$= \left(r \cos \theta , r \sin \theta\right)$

$= \left(- 64 \cos \left(\frac{5 \pi}{12}\right) , - 64 \sin \left(\frac{5 \pi}{12}\right)\right)$

$= \left(- 64 \cos \left({75}^{\circ}\right) , - 64 \sin \left({75}^{\circ}\right)\right)$

Now we can use the additive angle formulas

$\sin \left(u + v\right) = \sin u \sin v + \cos u \cos v$,
$\cos \left(u + v\right) = \cos u \cos v - \sin u \sin v$,

to get

$= \left(- 64 \cos \left({30}^{\circ} + {45}^{\circ}\right) , - 64 \sin \left({30}^{\circ} + {45}^{\circ}\right)\right)$

$= \left(- 64 \left(\cos {30}^{\circ} \cos {45}^{\circ} - \sin {30}^{\circ} \sin {45}^{\circ}\right) , - 64 \left(\sin {30}^{\circ} \cos {45}^{\circ} + \cos {30}^{\circ} \sin {45}^{\circ}\right)\right)$

$= \left(- 64 \left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2}\right) , - 64 \left(\frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}\right)\right)$

$= \left(- 64 \left(\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}\right) , - 64 \left(\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}\right)\right)$

$= \left(- 64 \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) , - 64 \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)\right)$

$= \left(- 16 \left(\sqrt{6} - \sqrt{2}\right) , - 16 \left(\sqrt{6} + \sqrt{2}\right)\right)$

$= \textcolor{b l u e}{\left(- 16 \sqrt{6} + 16 \sqrt{2} \text{,} - 16 \sqrt{6} - 16 \sqrt{2}\right)}$.