What is the distance between the following polar coordinates?: # (8,(7pi)/4), (5,(15pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (8,(7pi)/4), (5,(15pi)/8)   #

Christopher Agresta · Accepted Answer

The distance is #~~ 3.88#

The origin and these two points:

#(8, (7pi)/7) and (5, (15pi)/8)#

Form a triangle with,

side #a = 8#, side #b = 5#,

and the angle between them is,

#C = (15pi)/8 - (7pi)/4 = pi/8#.

The distance between the two points is the length of side c, in the equation for the Law of Cosines:

#c^2 = a^2 + b^2 - 2(a)(b)cos(C)#

#c^2 = 8^2 + 5^2 - 2(8)(5)cos(pi/8)#

#c ~~ 3.88#

Christopher Allers · Answer

undefined To find the distance between two polar coordinates, we can use the formula:

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

Given the coordinates:

$$r_1 = 8, \theta_1 = \frac{7\pi}{4}$$  
$$r_2 = 5, \theta_2 = \frac{15\pi}{8}$$

Plugging in these values, we get:

$$d = \sqrt{8^2 + 5^2 - 2*8*5*\cos\left(\frac{15\pi}{8} - \frac{7\pi}{4}\right)}$$

Simplifying the angle difference:

$$\frac{15\pi}{8} - \frac{7\pi}{4} = \frac{15\pi}{8} - \frac{14\pi}{8} = \frac{\pi}{8}$$

Plugging this back into the formula:

$$d = \sqrt{64 + 25 - 80\cos\left(\frac{\pi}{8}\right)}$$

Using the trigonometric identity:

$$\cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$$

Plugging this back into the formula:

$$d = \sqrt{64 + 25 - 80*\frac{\sqrt{2 + \sqrt{2}}}{2}}$$  
$$d = \sqrt{89 - 40\sqrt{2 + \sqrt{2}}}$$