What is the Cartesian form of #(12,(-7pi)/3))#?

Question

Charles Arocha · Accepted Answer

#(5 , -10.39 )#

It can be seen from the diagram that the x coordinate of P is #rcos(theta)# and the y coordinate of P is #rsin(theta)#

So:

#x=rcos(theta)#

#y=rsin(theta)#

From example:

#x=12cos((-7pi)/3)=-6#

#y=12sin((-7pi)/3)=-10.39# (2 .d.p.)

Cartesian coordinate:

#(5 , -10.39 )#

Elijah Abram · Answer

undefined The Cartesian form of the point $ (12, -\frac{7\pi}{3}) $ is $ (12, -\frac{7\pi}{3}) $. However, if you meant to convert the point from polar coordinates to Cartesian coordinates, the process is as follows:

Given the point $ (r, \theta) = (12, -\frac{7\pi}{3}) $, where $ r $ is the distance from the origin and $ \theta $ is the angle in radians measured counterclockwise from the positive x-axis, we can use the formulas:

$$ x = r \cos(\theta) $$
$$ y = r \sin(\theta) $$

Substituting the given values:

$$ x = 12 \cos\left(-\frac{7\pi}{3}\right) $$
$$ y = 12 \sin\left(-\frac{7\pi}{3}\right) $$

Now, simplify:

$$ x = 12 \cos\left(-\frac{2\pi}{3}\right) $$
$$ y = 12 \sin\left(-\frac{2\pi}{3}\right) $$

$$ x = 12 \left(-\frac{1}{2}\right) $$
$$ y = 12 \left(-\frac{\sqrt{3}}{2}\right) $$

$$ x = -6 $$
$$ y = -6\sqrt{3} $$

So, the Cartesian form of $ (12, -\frac{7\pi}{3}) $ is $ (-6, -6\sqrt{3}) $.