What is the distance between $(2, \frac{3 π}{4})$ $(2 ,(3 pi)/4 )$ and $(2, \frac{13 π}{8})$ $(2 , (13 pi )/8 )$ ?

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Answer 1 · 2016-09-20T14:33:16

$4 cos (\frac{π}{16}) \approx 4 (0.9808) = 3.9232$ $4cos(pi/16)~~4(0.9808)=3.9232$ .

The Distance btwn. pts. $P (r_{1}, θ_{1}) and Q (r_{2}, θ_{2})$ $P(r_1,theta_1) and Q(r_2,theta_2)$ , i.e.,

$P Q$ $PQ$ , is given by,

$P Q^{2} = r_{1}^{2} + r_{2}^{2} - 2 r_{1} \cdot r_{2} \cdot cos (θ_{1} - θ_{2})$ $PQ^2=r_1^2+r_2^2-2r_1*r_2*cos(theta_1-theta_2)$ .

In our case, if $d$ $d$ is the reqd. dist., then,

$d^{2} = 2^{2} + 2^{2} - 2 \cdot 2 \cdot 2 cos (13 \frac{π}{8} - 3 \frac{π}{4})$ $d^2=2^2+2^2-2*2*2cos(13pi/8-3pi/4)$

$= 8 - 8 cos (7 \frac{π}{8})$ $=8-8cos(7pi/8)$

$= 8 - 8 cos (π - \frac{π}{8})$ $=8-8cos(pi-pi/8)$

$= 8 + 8 cos (\frac{π}{8})$ $=8+8cos(pi/8)$

$= 8 (1 + cos (\frac{π}{8}))$ $=8(1+cos(pi/8))$

$= 8 \cdot 2 {cos}^{2} (\frac{1}{2} \cdot \frac{π}{8}) = 16 {cos}^{2} (\frac{π}{16})$ $=8*2cos^2(1/2*pi/8)=16cos^2(pi/16)$ .

$:. d=4cos(pi/16)$ .

Numerically, $d~~4(0.9808)=3.9232$ .

What is the distance between (2 ,(3 pi)/4 )(2,3π4)(2 ,(3 pi)/4 ) and (2 , (13 pi )/8 )(2,13π8)(2 , (13 pi )/8 )?