# What is the Cartesian form of (9,(1pi )/12)?

Apr 27, 2018

P(9,{1π}/12) in rectangular form is:

$\left(9 \cos \left(\frac{\pi}{12}\right) , 9 \sin \left(\frac{\pi}{12}\right)\right) = \left(\frac{9}{4} \left(\sqrt{6} + \sqrt{2}\right) , \frac{9}{4} \left(\sqrt{6} - \sqrt{2}\right)\right)$

#### Explanation:

$\left(x , y\right) = \left(9 \cos \left(\frac{\pi}{12}\right) , 9 \sin \left(\frac{\pi}{12}\right)\right)$

$\frac{\pi}{12} = {180}^{\circ} / 12 = {15}^{\circ}$

We can get the trig functions of ${15}^{\circ}$ from the half angle formula or the difference angle formula.

$\cos {15}^{\circ} = \cos \left(\frac{1}{2} \left({30}^{\circ}\right)\right) = \cos \left({45}^{\circ} - {30}^{\circ}\right)$

The difference angle form avoids nested square roots; let's use that.

cos 15^circ = cos 45^circ cos 30^circ + sin 45^circ sin 30^circ

cos 15^circ = (sqrt{2}/2)(sqrt{3}/2) + (sqrt{2}/2)(1/2)

$\cos {15}^{\circ} = \frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right)$

Similarly,

$\sin {15}^{\circ} = \sin {45}^{\circ} \cos {30}^{\circ} - \cos {45}^{\circ} \sin {30}^{\circ}$

$\sin {15}^{\circ} = \frac{1}{4} \left(\setminus \sqrt{6} - \sqrt{2}\right)$

Putting it together, P(9,{1π}/12) in rectangular form is:

$\left(9 \cos \left(\frac{\pi}{12}\right) , 9 \sin \left(\frac{\pi}{12}\right)\right) = \left(\frac{9}{4} \left(\sqrt{6} + \sqrt{2}\right) , \frac{9}{4} \left(\sqrt{6} - \sqrt{2}\right)\right)$