What is the distance the polar coordinates #(-2 ,( -3 )/8 )# and #(6 ,(-7 pi )/4 )#?

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Answer 1 · 2017-01-10T09:49:16

#=sqrt(40-24sqrt(((sqrt2-1)/(2sqrt2))))5.551#, nearly.

After correcting the first angle as #-3/8pi#, the two points are

P(-2, -3/8pi) or, with positive r, P(2. -3/8pi-pi) =#

# P(2, -11/8pi) and Q(6, -7/4pi)#

Now, in the #triangle POQ#, where O is the pole r = 0,

#OP = 2, OQ = 6 and angle POQ = (--7/4pi-(-11/8pi))=-3/8pi=-67.5^o#.

Use #PQ=sqrt(OP^2+OQ^2-2 OP OQ cos angle POQ#

#=sqrt(4+36-24cos(-67.5^o)#

Here, #cos(-67.5^0)=cos(67.5^o)#

#=sin (90-67.5)^o=sin 22.5^o#

#=sqrt((1-cos 45^0)/2)=sqrt(((sqrt2-1)/(2sqrt2)) #. So,

#PQ=sqrt(40-24sqrt(((sqrt2-1)/(2sqrt2))))=5.551#, nearly -