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 -

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