What is the distance between the following polar coordinates?: # (4,(-11pi)/12), (1,(-7pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (4,(-11pi)/12), (1,(-7pi)/8)   #

Luke Alvarez · Accepted Answer

#P_1P_2=sqrt(9.068441109) ~~3.01# Distance formula in Polar form:#P_1P_2=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_2-theta_1))# #r_1=4, theta_1=(-11pi)/12, r_2=1,theta_2=(-7pi)/8# #P_1P_2=sqrt(4^2+1^2-2(4)(1)cos((-7pi)/8-(-11pi)/12))# #P_1P_2=sqrt(16+1-8cos((pi)/24))# #P_1P_2=sqrt(17-8cos((pi)/24))# #P_1P_2=sqrt(9.068441109) ~~3.01#

Paisley Johnson · Answer

undefined To find the distance between two polar coordinates $(r_1, \theta_1)$ and $(r_2, \theta_2)$, you can use the formula:

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

Given the polar coordinates $(4, -\frac{11\pi}{12})$ and $(1, -\frac{7\pi}{8})$, we have:

$$r_1 = 4, \quad \theta_1 = -\frac{11\pi}{12}$$
$$r_2 = 1, \quad \theta_2 = -\frac{7\pi}{8}$$

Plugging these values into the formula:

$$d = \sqrt{(4)^2 + (1)^2 - 2(4)(1) \cos\left(-\frac{7\pi}{8} - \left(-\frac{11\pi}{12}\right)\right)}$$

$$d = \sqrt{16 + 1 - 8\cos\left(-\frac{7\pi}{8} + \frac{11\pi}{12}\right)}$$

$$d = \sqrt{17 - 8\cos\left(\frac{11\pi}{12} - \frac{7\pi}{8}\right)}$$

$$d = \sqrt{17 - 8\cos\left(\frac{11\pi}{12} - \frac{21\pi}{24}\right)}$$

$$d = \sqrt{17 - 8\cos\left(\frac{\pi}{24}\right)}$$

Since $\cos\left(\frac{\pi}{24}\right)$ is a known value, we can substitute it and compute the distance:

$$d = \sqrt{17 - 8\cos\left(\frac{\pi}{24}\right)}$$
$$d = \sqrt{17 - 8\sqrt{\frac{1 + \sqrt{3}}{8}}}$$

$$d \approx \sqrt{17 - 4\sqrt{1 + \sqrt{3}}}$$

Therefore, the distance between the given polar coordinates is approximately $\sqrt{17 - 4\sqrt{1 + \sqrt{3}}}$.