What is the distance between #(2 ,(7 pi)/6 )# and #(3 , (- pi )/8 )#?

1 Answer
Jan 3, 2016

#1.0149#

Explanation:

The distance formula for polar coordinates is

#d=sqrt(r_1^2+r_2^2-2r_1r_2Cos(theta_1-theta_2)#
Where #d# is the distance between the two points, #r_1#, and #theta_1# are the polar coordinates of one point and #r_2# and #theta_2# are the polar coordinates of another point.
Let #(r_1,theta_1)# represent #(2,(7pi)/6)# and #(r_2,theta_2)# represent #(3,-pi/8)#.
#implies d=sqrt(2^2+3^2-2*2*3Cos((7pi)/6-(-pi/8))#
#implies d=sqrt(4+9-12Cos((7pi)/6+pi/8)#
#implies d=sqrt(13-12cos((28pi+3pi)/24))=sqrt(13-12cos((31pi)/24))=sqrt(13-12cos(4.0558))=sqrt(13-12*0.9975)=sqrt(13-12*0.9975)=sqrt(13-11.97)=sqrt(1.03)=1.0149# units
#implies d=1.0149# units (approx)
Hence the distance between the given points is #1.0149#.