# How do you convert r = 3 sec ( pi/3 - theta)  into cartesian form?

Oct 12, 2016

Please see the explanation for the conversion process.

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

#### Explanation:

Multiply both sides of the equation by $\cos \left(\frac{\pi}{3} - \theta\right)$ (because $\cos \left(A\right) \sec \left(A\right) = 1$):

$r \cos \left(\frac{\pi}{3} - \theta\right) = 3$

Use the identity $\cos \left(A - B\right) = \cos \left(A\right) \cos \left(B\right) + \sin \left(A\right) \sin \left(B\right)$:

$r \left(\cos \left(\frac{\pi}{3}\right) \cos \left(\theta\right) + \sin \left(\frac{\pi}{3}\right) \sin \left(\theta\right)\right) = 3$

We know the values for the sine and cosine of $\frac{\pi}{3}$:

$r \left(\frac{1}{2} \cos \left(\theta\right) + \frac{\sqrt{3}}{2} \sin \left(\theta\right)\right) = 3$

Distribute r:

$\frac{1}{2} r \cos \left(\theta\right) + \frac{\sqrt{3}}{2} r \sin \left(\theta\right) = 3$

Substitute y for $r \sin \left(\theta\right)$ and x for $r \cos \left(\theta\right)$

$\frac{1}{2} x + \frac{\sqrt{3}}{2} y = 3$

$\frac{\sqrt{3}}{2} y = - \frac{1}{2} x + 3$

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