What is the distance between the following polar coordinates?: # (2,(pi)/2), (2,(17pi)/12) #

Question

What is the distance between the following polar coordinates?: #   (2,(pi)/2), (2,(17pi)/12)   #

Paisley Johnson · Accepted Answer

#bar(AB)\approx3.9658#

Graph the two points:

A triangle is formed when the two points #A# and #B# are connected to the origin #O#. We know the angle #/_AOB# is #(17pi)/12-pi/2=(11pi)/12#, and the two sides #bar(AO)=bar(BO)=2#.

Then, #bar(AB)# can be found using the law of cosines:
#bar(AB)^2=bar(AO)^2+bar(BO)^2-2*bar(AO)*bar(BO)*cos(/_AOB)#
#bar(AB)^2=2^2+2^2-2*2*2*cos((11pi)/12)#
#bar(AB)^2\approx15.7274#

#thereforebar(AB)\approx3.9658#

Cameron Alexander · Answer

undefined To find the distance between the polar coordinates (2, π/2) and (2, 17π/12), you can use the polar distance formula:

$$ d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_2 - \theta_1)} $$

Where $ r_1 $ and $ r_2 $ are the radii of the polar coordinates, and $ \theta_1 $ and $ \theta_2 $ are the angles.

Given $ r_1 = r_2 = 2 $ and $ \theta_1 = \frac{\pi}{2} $ and $ \theta_2 = \frac{17\pi}{12} $, plug these values into the formula:

$$ d = \sqrt{2^2 + 2^2 - 2(2)(2)\cos\left(\frac{17\pi}{12} - \frac{\pi}{2}\right)} $$

$$ d = \sqrt{4 + 4 - 8\cos\left(\frac{17\pi}{12} - \frac{\pi}{2}\right)} $$

$$ d = \sqrt{8 - 8\cos\left(\frac{17\pi}{12} - \frac{\pi}{2}\right)} $$

To evaluate $ \cos\left(\frac{17\pi}{12} - \frac{\pi}{2}\right) $, first find $ \frac{17\pi}{12} - \frac{\pi}{2} $:

$$ \frac{17\pi}{12} - \frac{\pi}{2} = \frac{17\pi - 6\pi}{12} - \frac{6\pi}{12} = \frac{11\pi}{12} $$

Now, evaluate $ \cos\left(\frac{11\pi}{12}\right) $.

Using trigonometric identities, you'll find that $ \cos\left(\frac{11\pi}{12}\right) = -\frac{\sqrt{2 + \sqrt{3}}}{2} $.

Substitute this value back into the equation:

$$ d = \sqrt{8 - 8\left(-\frac{\sqrt{2 + \sqrt{3}}}{2}\right)} $$

$$ d = \sqrt{8 + 4\sqrt{2 + \sqrt{3}}} $$

Thus, the distance between the polar coordinates is $ \sqrt{8 + 4\sqrt{2 + \sqrt{3}}} $.