# How do you convert Theta=-pi/6 into cartesian form?

Jun 3, 2016

I adhere to the convention that r is modulus (length), and so, r is non-negative. $y = - \frac{x}{\sqrt{3}} , y < 0$.

#### Explanation:

Use theta = tan^(-1)(y/x)#

Here, ${\tan}^{- 1} \left(\frac{y}{x}\right) = \theta = - \frac{\pi}{6}$.

So,$\frac{y}{x} = \tan \left(- \frac{\pi}{6}\right) = - \tan \left(\frac{\pi}{6}\right) = - \frac{1}{\sqrt{3}}$.

This line is restricted to the 4th quadrant.

The other half is given by the polar equation $\theta = \left(\frac{5}{6}\right) \pi$.

Note that $\tan \left(\left(\frac{5}{6}\right) \pi\right) = - \frac{1}{\sqrt{3}}$.

You can include r = 0 ((x, y)=(0, 0)) as the limiting end-point, for each

half.