# What is the distance between the following polar coordinates?:  (3,(23pi)/12), (7,(13pi)/8)

Apr 13, 2016

The distance between the two points is $5.7$

#### Explanation:

It's easier to calculate with cartesian coordinates than with polar, so my advice is to start with converting (3;(23pi)/12) and (7;(13pi)/8) to (x_1;y_1) and (x_2;y_2)

The relationship between $\textcolor{b l u e}{c a r t e s i a n}$ and $\textcolor{red}{p o l a r}$ coordinates is:

color(blue)(x)=color(red)(R cdot cos(theta)
color(blue)(y)=color(red)(R cdot sin(theta)

For your first point:
$R = 3$
$\theta = \frac{23 \pi}{12}$

For your second point:
$R = 7$
$\theta = \frac{13 \pi}{8}$

The distance between two points is:
$d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

$= \sqrt{{\left(3 \cos \left(\frac{23 \pi}{12}\right) - 7 \cos \left(\frac{13 \pi}{8}\right)\right)}^{2} + {\left(3 \sin \left(\frac{23 \pi}{12}\right) - 7 \sin \left(\frac{13 \pi}{8}\right)\right)}^{2}}$

$= \sqrt{{\left(3 \cos \left(- \frac{\pi}{12}\right) - 7 \cos \left(- \frac{3 \pi}{8}\right)\right)}^{2} + {\left(3 \sin \left(- \frac{\pi}{12}\right) - 7 \sin \left(- \frac{3 \pi}{8}\right)\right)}^{2}}$

$= \sqrt{{\left(3 \cos \left(\frac{\pi}{12}\right) - 7 \cos \left(\frac{3 \pi}{8}\right)\right)}^{2} + {\left(7 \sin \left(\frac{3 \pi}{8}\right) - 3 \sin \left(\frac{\pi}{12}\right)\right)}^{2}}$

$\approx 5.7$