# What is the Cartesian form of (36,(-pi)/16))?

Sep 22, 2017

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

$= \left(18 \sqrt{2 + \sqrt{2 + \sqrt{2}}} , - 18 \sqrt{2 - \sqrt{2 + \sqrt{2}}}\right)$

#### Explanation:

To convert from polar to rectangular coordinates use:

$\left\{\begin{matrix}x = r \cos \theta \\ y = r \sin \theta\end{matrix}\right.$

First let's find algebraic expressions for $\sin \left(\frac{\pi}{16}\right)$ and $\cos \left(\frac{\pi}{16}\right)$

Use:

$\sin \left(\frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

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

$\cos \left(2 \theta\right) = {\cos}^{2} \theta - {\sin}^{2} \theta = 2 {\cos}^{2} \theta - 1 = 1 - 2 {\sin}^{2} \theta$

$\sin \left(- \theta\right) = - \sin \left(\theta\right)$

$\cos \left(- \theta\right) = \cos \left(\theta\right)$

From $\cos 2 \theta = 2 {\cos}^{2} \theta - 1$ we can deduce:

$\cos \theta = \pm \sqrt{\frac{\cos 2 \theta + 1}{2}} = \pm \frac{1}{2} \sqrt{2 + 2 \cos 2 \theta}$

From $\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$ we can deduce:

$\sin \theta = \pm \sqrt{\frac{1 - \cos 2 \theta}{2}} = \pm \frac{1}{2} \sqrt{2 - 2 \cos 2 \theta}$

Since $\frac{\pi}{8}$ and $\frac{\pi}{16}$ are both in Q1, we know that $\sin$ and $\cos$ are both positive, so we can use the positive square root each time to find:

$\sin \left(\frac{\pi}{16}\right) = \frac{1}{2} \sqrt{2 - 2 \cos \left(\frac{\pi}{8}\right)}$

$\textcolor{w h i t e}{\sin \left(\frac{\pi}{16}\right)} = \frac{1}{2} \sqrt{2 - \sqrt{2 + 2 \cos \left(\frac{\pi}{4}\right)}}$

$\textcolor{w h i t e}{\sin \left(\frac{\pi}{16}\right)} = \frac{1}{2} \sqrt{2 - \sqrt{2 + \sqrt{2}}}$

$\cos \left(\frac{\pi}{16}\right) = \frac{1}{2} \sqrt{2 + 2 \cos \left(\frac{\pi}{8}\right)}$

$\textcolor{w h i t e}{\cos \left(\frac{\pi}{16}\right)} = \frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2}}}$

So:

$\sin \left(- \frac{\pi}{16}\right) = - \sin \left(\frac{\pi}{16}\right) = - \frac{1}{2} \sqrt{2 - \sqrt{2 + \sqrt{2}}}$

$\cos \left(- \frac{\pi}{16}\right) = \cos \left(\frac{\pi}{16}\right) = \frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2}}}$

So with $r = 36$ and $\theta = - \frac{\pi}{16}$ we find:

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

$\textcolor{w h i t e}{\left(x , y\right)} = \left(36 \left(\frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2}}}\right) , 36 \left(- \frac{1}{2} \sqrt{2 - \sqrt{2 + \sqrt{2}}}\right)\right)$

$\textcolor{w h i t e}{\left(x , y\right)} = \left(18 \sqrt{2 + \sqrt{2 + \sqrt{2}}} , - 18 \sqrt{2 - \sqrt{2 + \sqrt{2}}}\right)$