# What is the distance between (1 ,(3 pi)/4 ) and (3 , (15 pi )/8 )?

Mar 10, 2016

Distance between $\left(1 , \frac{3 \pi}{4}\right)$ and $\left(3 , \frac{15 \pi}{8}\right)$ is $3.9425$

#### Explanation:

$\left(r , \theta\right)$ in polar coordinates is $\left(r \cos \theta , r \sin \theta\right)$ in rectangular coordinates.

Hence, $\left(1 , \frac{3 \pi}{4}\right)$ in rectangular coordinates is $\left(\cos \left(\frac{3 \pi}{4}\right) , \sin \left(\frac{3 \pi}{4}\right)\right)$ or $\left(- \frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{2}\right)$ or $\left(- 0.7071 , 0.7071\right)$

And $\left(3 , \frac{15 \pi}{8}\right)$ in rectangular coordinates is $\left(3 \cos \left(\frac{15 \pi}{8}\right) , 3 \sin \left(\frac{15 \pi}{8}\right)\right)$ or $\left(2.7716 , - 1.1481\right)$

Hence distance between $\left(- 0.7071 , 0.7071\right)$ and $\left(2.7716 , - 1.1481\right)$ is

$\sqrt{{\left(2.7716 + 0.7071\right)}^{2} + {\left(- 1.1481 - 0.7071\right)}^{2}}$ or
$\sqrt{{\left(3.4787\right)}^{2} + {\left(- 1.8552\right)}^{2}}$ or
$\sqrt{12.1014 + 3.4418} = \sqrt{15.5432} = 3.9425$