# What is the Cartesian form of (71,(13pi) /12)?

$\left(- \frac{71 \cdot \sqrt{2 + \sqrt{3}}}{2} , - \frac{71 \cdot \sqrt{2 - \sqrt{3}}}{2}\right)$

#### Explanation:

$x = r \cdot \cos \theta$ and $y = r \cdot \sin \theta$

Note: that $\frac{\pi}{12} = \frac{\frac{\pi}{6}}{2}$

Use half angle formulas
$\sin \left(\frac{A}{2}\right) = \sqrt{\frac{1 - \cos A}{2}}$

and $\cos \left(\frac{A}{2}\right) = \sqrt{\frac{1 + \cos A}{2}}$

$x = 71 \cdot \cos \left(\frac{13 \pi}{12}\right)$

$x = 71 \cdot \left(- \frac{\sqrt{2 + \sqrt{3}}}{2}\right)$

$x = - \frac{71 \cdot \sqrt{2 + \sqrt{3}}}{2}$

$y = 71 \cdot \sin \left(\frac{13 \pi}{12}\right)$

$y = 71 \cdot \left(- \frac{\sqrt{2 - \sqrt{3}}}{2}\right)$

$y = - \frac{71 \cdot \sqrt{2 - \sqrt{3}}}{2}$