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

Aug 2, 2018

#### Answer:

$D = \sqrt{17 - 8 \cos \left(\frac{\pi}{8}\right)}$

#### Explanation:

We know that ,

$\text{Distance between Polar Co-ordinates:} A \left({r}_{1} , {\theta}_{1}\right) \mathmr{and} B \left({r}_{2} , {\theta}_{2}\right)$ is

color(red)(D=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_1-theta_2))...to(I)

We have , ${P}_{1} \left(4 , \frac{- 3 \pi}{8}\right) \mathmr{and} {P}_{2} \left(- 1 , \frac{3 \pi}{4}\right)$.

So , ${r}_{1} = 4 , {r}_{2} = - 1 , {\theta}_{1} = \frac{- 3 \pi}{8} \mathmr{and} {\theta}_{2} = \frac{3 \pi}{4}$

$\implies {\theta}_{1} - {\theta}_{2} = \frac{- 3 \pi}{8} - \frac{3 \pi}{4} = \frac{- 3 \pi - 6 \pi}{8} = \frac{- 9 \pi}{8}$

$\implies \cos \left({\theta}_{1} - {\theta}_{2}\right) = \cos \left(\frac{- 9 \pi}{8}\right)$
$\implies \cos \left({\theta}_{1} - {\theta}_{2}\right) = \cos \left(\frac{9 \pi}{8}\right) \to \left[\because \cos \left(- \theta\right) = \cos \theta\right]$
$\implies \cos \left({\theta}_{1} - {\theta}_{2}\right) = \cos \left(\pi + \frac{\pi}{8}\right) = - \cos \left(\frac{\pi}{8}\right)$

"Using : " color(red)((I) we get

D=sqrt(4^2+(-1)^2-2(4)(-1)(-cos(pi/8))

$\implies D = \sqrt{16 + 1 - 8 \cdot \cos \left(\frac{\pi}{8}\right)}$

$\implies D = \sqrt{17 - 8 \cos \left(\frac{\pi}{8}\right)}$

$\implies D \approx 3$