What is the distance between the following polar coordinates?: # (6,(7pi)/12), (2,(-3pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (6,(7pi)/12), (2,(-3pi)/8)   #

Ezra Adams · Accepted Answer

#d = 2sqrt(10 - 3sqrt( 2 + sqrt ( 2 - sqrt 3 ) )# #d = sqrt(r_1^2+r_2^2 - 2r_1r_2cos(phi_2-phi_1))# #d = sqrt(6^2+2^2 - 2*6*2*cos(-(3pi)/8-(7pi)/12)# #d = sqrt(40 - 24*cos((-29pi)/24)# #d = 2sqrt(10 - 6cos((29pi)/24)# #cos((29pi)/6) = cos(4pi + (5pi)/6) = cos((5pi)/6) = -sqrt3/2# #cos((29pi)/12) = cos(((29pi)/6)/2) = sqrt((1+cos((29pi)/6))/2)# #cos((29pi)/12) = sqrt((1-sqrt3/2)/2) = sqrt((2-sqrt3)/4) = sqrt(2-sqrt3) /2 # #cos((29pi)/24) = cos(((29pi)/12)/2) = sqrt((1+cos((29pi)/12))/2)# #cos((29pi)/24) = sqrt( ( 1+ sqrt( ( 2-sqrt3 ) ) / 2 ) /2 ) # #cos((29pi)/24) = sqrt( 2 + sqrt ( 2 - sqrt 3 ) ) /2# #d = 2sqrt(10 - 6* sqrt( 2 + sqrt ( 2 - sqrt 3 ) ) /2)# #d = 2sqrt(10 - 3sqrt( 2 + sqrt ( 2 - sqrt 3 ) )#

Cameron Alexander · Answer

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

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

Plugging in the given values:

$$ r_1 = 6, \theta_1 = \frac{7\pi}{12}, r_2 = 2, \theta_2 = -\frac{3\pi}{8} $$

$$ \text{Distance} = \sqrt{(6^2 + 2^2 - 2(6)(2)\cos\left(\frac{7\pi}{12} - (-\frac{3\pi}{8})\right))} $$

$$ = \sqrt{(36 + 4 - 24\cos(\frac{7\pi}{12} + \frac{3\pi}{8}))} $$

$$ = \sqrt{(40 - 24\cos(\frac{7\pi}{12} + \frac{3\pi}{8}))} $$

Since $\cos(\frac{7\pi}{12} + \frac{3\pi}{8})$ cannot be simplified further without using trigonometric identities, we'll leave it in this form. Thus, the distance between the given polar coordinates is:

$$ \text{Distance} = \sqrt{(40 - 24\cos(\frac{7\pi}{12} + \frac{3\pi}{8}))} $$