What is the distance between $(4, \frac{- π}{8})$ $(4 ,( - pi)/8 )$ and $(- 1, \frac{3 π}{2})$ $(-1 ,( 3 pi )/2 )$ ?

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What is the distance between $(4, \frac{- π}{8})$ $(4 ,( - pi)/8 )$ and $(- 1, \frac{3 π}{2})$ $(-1 ,( 3 pi )/2 )$ ?

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Answer 1 · 2016-04-15T17:31:33

$P_{1} P_{2} \approx 4.47$ $P_1P_2~~4.47$

Explanation:

Use formula $P_{1} P_{2} = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} cos (θ_{2} - θ_{1})}$ $P_1P_2=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_2-theta_1))$

$r_{1} = 4, θ_{1} = - \frac{π}{8} and r_{2} = - 1, θ_{2} = \frac{3 π}{2}$ $r_1=4,theta_1=-pi/8 and r_2=-1,theta_2=(3pi)/2$

$P_{1} P_{2} = \sqrt{4^{2} + {(- 1)}^{2} - 2 (4) (- 1) cos ((\frac{3 π}{2}) - \frac{- π}{8})}$ $P_1P_2=sqrt(4^2+(-1)^2-2(4)(-1)cos(((3pi)/2)-(-pi)/8)$

$= \sqrt{16 + 1 + 8 cos (\frac{13 π}{8})}$ $=sqrt(16+1+8cos((13pi)/8)$

$= \sqrt{17 + 8 cos (\frac{13 π}{8})}$ $=sqrt(17+8cos((13pi)/8))$

$P_{1} P_{2} \approx 4.47$ $P_1P_2~~4.47$

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What is the distance between $(4, \frac{- π}{8})$ $(4 ,( - pi)/8 )$ and $(- 1, \frac{3 π}{2})$ $(-1 ,( 3 pi )/2 )$ ?

1 Answer

Explanation:

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