# What is the Cartesian form of (45,(-13pi)/8)?

Sep 22, 2017

$\left(45 \cos \left(- \frac{13 \pi}{8}\right) , 45 \sin \left(- \frac{13 \pi}{8}\right)\right) = \left(\frac{45}{2} \sqrt{2 - \sqrt{2}} , \frac{45}{2} \sqrt{2 + \sqrt{2}}\right)$

#### Explanation:

The cartesian form of $\left(r , \theta\right)$ is $\left(r \cos \theta , r \sin \theta\right)$

Let's find $\cos \left(- \frac{13 \pi}{8}\right)$ and $\sin \left(- \frac{13 \pi}{8}\right)$ first.

Note that $- \frac{13 \pi}{4}$ is coterminal with $\frac{3 \pi}{4}$ in Q2.

Hence:

$\left\{\begin{matrix}\cos \left(- \frac{13 \pi}{4}\right) = - \frac{\sqrt{2}}{2} \\ \sin \left(- \frac{13 \pi}{4}\right) = \frac{\sqrt{2}}{2}\end{matrix}\right.$

Then note that:

$\cos 2 \theta = 2 {\cos}^{2} \theta - 1 = 1 - 2 {\sin}^{2} \theta$

Hence:

$\cos \left(- \frac{13 \pi}{8}\right) = \pm \sqrt{\frac{1 + \cos \left(- \frac{13 \pi}{4}\right)}{2}}$

$\textcolor{w h i t e}{\cos \left(- \frac{13 \pi}{8}\right)} = \pm \frac{1}{2} \sqrt{2 + 2 \cos \left(- \frac{13 \pi}{4}\right)}$

$\textcolor{w h i t e}{\cos \left(- \frac{13 \pi}{8}\right)} = \pm \frac{1}{2} \sqrt{2 - \sqrt{2}}$

$\sin \left(- \frac{13 \pi}{8}\right) = \pm \sqrt{\frac{1 - \cos \left(- \frac{13 \pi}{4}\right)}{2}}$

$\textcolor{w h i t e}{\sin \left(- \frac{13 \pi}{8}\right)} = \pm \frac{1}{2} \sqrt{2 - 2 \cos \left(- \frac{13 \pi}{4}\right)}$

$\textcolor{w h i t e}{\sin \left(- \frac{13 \pi}{8}\right)} = \pm \frac{1}{2} \sqrt{2 + \sqrt{2}}$

Which signs are correct?

Note that $- \frac{13 \pi}{8}$ is coterminal with $\frac{3 \pi}{8}$ which is in Q1.

So $\cos$ and $\sin$ are both positive and:

$\left\{\begin{matrix}\cos \left(- \frac{13 \pi}{8}\right) = \frac{1}{2} \sqrt{2 - \sqrt{2}} \\ \sin \left(- \frac{13 \pi}{8}\right) = \frac{1}{2} \sqrt{2 + \sqrt{2}}\end{matrix}\right.$

So polar coordinates $\left(45 , - \frac{13 \pi}{8}\right)$ in rectangular coordinates is:

$\left(45 \cos \left(- \frac{13 \pi}{8}\right) , 45 \sin \left(- \frac{13 \pi}{8}\right)\right) = \left(\frac{45}{2} \sqrt{2 - \sqrt{2}} , \frac{45}{2} \sqrt{2 + \sqrt{2}}\right)$