What is the Cartesian form of #(-7,(11pi)/8)#?

Question

Emily Ammerman · Accepted Answer

Cartesian form of coordinates are #(2.6789,6.4673)# Cartesian form of #(r,theta)# is #(rcostheta,rsintheta)# Hence for #(-7, 11pi/8)#, it will be #(-7cos(11pi/8), -7sin(11pi/8))# As #cos(11pi/8)=-0.3827# and #sin(11pi/8)=-0.9239# Hence cartesian form of coordinates are #(2.6789,6.4673)#

Axel Alvarez · Answer

undefined The Cartesian form of the point $(-7, \frac{11\pi}{8})$ can be found by converting the polar coordinates to rectangular coordinates using the following formulas:

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

where $ r $ is the radius or distance from the origin, and $ \theta $ is the angle measured counterclockwise from the positive x-axis.

Given that $ r = -7 $ and $ \theta = \frac{11\pi}{8} $, we can plug these values into the formulas to find the rectangular coordinates:

$$ x = -7 \cdot \cos\left(\frac{11\pi}{8}\right) $$
$$ y = -7 \cdot \sin\left(\frac{11\pi}{8}\right) $$

Computing the values:

$$ x = -7 \cdot \cos\left(\frac{11\pi}{8}\right) \approx -7 \cdot \cos\left(\frac{3\pi}{8}\right) $$
$$ x \approx -7 \cdot \left(-\frac{\sqrt{2+\sqrt{2}}}{2}\right) $$
$$ x \approx -7 \cdot \left(-\frac{\sqrt{2}+\sqrt{2}}{2}\right) $$
$$ x \approx -7 \cdot \left(-\frac{2\sqrt{2}}{2}\right) $$
$$ x \approx 7\sqrt{2} $$

$$ y = -7 \cdot \sin\left(\frac{11\pi}{8}\right) \approx -7 \cdot \sin\left(\frac{3\pi}{8}\right) $$
$$ y \approx -7 \cdot \left(\frac{\sqrt{2-\sqrt{2}}}{2}\right) $$
$$ y \approx -7 \cdot \left(\frac{\sqrt{2}-\sqrt{2}}{2}\right) $$
$$ y \approx -7\sqrt{2} $$

So, the Cartesian form of the point $(-7, \frac{11\pi}{8})$ is $(7\sqrt{2}, -7\sqrt{2})$.