What is the distance between the following polar coordinates?: # (2,(3pi)/4), (1,(15pi)/8) #

3 Answers
Jul 18, 2018

#D=sqrt(5+4cos(pi/8))~~2.9488#

Explanation:

We know that ,

#"Distance between Polar Co-ordinates:"A(r_1,theta_1)and B(r_2,theta_2) # is

#color(red)(D=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_1-theta_2))...to(I)#

We have , #P_1(2,(3pi)/4) and P_2(1,(15pi)/8)#.

So , #r_1=2 , r_2=1 , theta_1=(3pi)/4 and theta_2=(15pi)/8#

#=>theta_1-theta_2=(3pi)/4-(15pi)/8=(6pi-15pi)/8=(-9pi)/8#

#=>cos(theta_1-theta_2)=cos((-9pi)/8)#
#=>cos(theta_1-theta_2)=cos((9pi)/8)to[becausecos(-theta)=costheta]#
#=>cos(theta_1-theta_2)=cos(pi+pi/8)=-cos(pi/8)#

#"Using : " color(red)((I)# we get

#D=sqrt(2^2+1^2-2(2)(1)(-cos(pi/8))#

#=>D=sqrt(4+1+4*cos(pi/8))#

#=>D=sqrt(5+4cos(pi/8))#

#=>D~~2.9488#

Jul 18, 2018

#sqrt(2+sqrt(2-sqrt(2)))#

Explanation:

It is nontrivial to determine polar distances (since drawing a line in polar coordinates isn't very easy), so we should convert to Cartesian coordinates.

The first point is moderately easy:
#x = rcostheta = 2 * -sqrt(2)/2 = -sqrt(2)#
#y = rsintheta = 2 * sqrt(2)/2 = sqrt(2)#

The second point is a little harder, but we only need to use half angle identities. Let #theta = (15pi)/4 equiv -(pi)/4#. Therefore, #cos(theta) = sqrt(2)/2# and

#cos(theta/2) = pm sqrt((1+costheta)/2) = pm sqrt(2+sqrt2)/2#
#sin(theta/2) = pm sqrt(1-costheta)/2 = pm sqrt(2-sqrt2)/2#

Since #(15pi)/8# is in the 4th quadrant, its cosine is positive and its sine is negative. Therefore,

#x = r cos((15pi)/8) = sqrt(2+sqrt2)/2#
#y = r sin((15pi)/8) = sqrt(2-sqrt2)/2#

This allows us to use Pythagorean theorem in order to find the distance between the two points:

#Delta x = sqrt(2+sqrt2)/2 + sqrt2#
#Delta y = sqrt(2-sqrt2)/2 - sqrt2#

Pulling out a factor of #sqrt(2)# from each, then squaring and adding them,
#ds^2 = 2[1 + sqrt2/2 + 1 + 1 - sqrt2/2 + 1 + 2sqrt(1+sqrt2/2) - 2 sqrt(1 - sqrt2/2)] #
#= 2[4 + 2(sqrt(1+sqrt2/2) - sqrt(1-sqrt2/2))] #
Using the cosine sum formula with #pi/8# and #pi/4#, we get
#= 8 (1 + cos ((3pi)/8) ) = 4 cos^2((3pi)/16)#

Therefore the total distance is
#D = 2 cos((3pi)/16) = sqrt(2 + sqrt(2 - sqrt(2))) approx 1.66#

#2.949\ \text{unit}#

Explanation:

The distance between the points #(r_1, \theta_1)\equiv(2, {3\pi}/4)# & #(r_2, \theta_2)\equiv(1, {15\pi}/8)# is given by the formula as follows

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

#=\sqrt{2^2+1^2-2(2)(1)\cos({3\pi}/4-{15\pi}/8)}#

#=\sqrt{5-4\cos({9\pi}/8)}#

#=\sqrt{5+4\cos({\pi}/8)}#

#=\sqrt{\frac{7\sqrt2+2}{\sqrt2}}#

#=2.949\ \text{unit}#