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What is the distance between #(4 ,( - 3pi)/8 )# and #(-1 ,( 3 pi )/4 )#?

1 Answer
Aug 2, 2018

Answer:

#D=sqrt(17-8cos(pi/8))#

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(4,(-3pi)/8) and P_2(-1,(3pi)/4)#.

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

#=>theta_1-theta_2=(-3pi)/8-(3pi)/4=(-3pi-6pi)/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(4^2+(-1)^2-2(4)(-1)(-cos(pi/8))#

#=>D=sqrt(16+1-8*cos(pi/8))#

#=>D=sqrt(17-8cos(pi/8))#

#=>D~~3#