# What is the distance between the following polar coordinates?:  (6,(19pi)/12), (3,(15pi)/8)

$4.804642396$

#### Explanation:

In general, the distance $d$ between the points $\left({r}_{1} , \setminus {\theta}_{1}\right)$ & $\left({r}_{2} , \setminus {\theta}_{2}\right)$ is given as follows

$d = \setminus \sqrt{{\left({r}_{1} \setminus \cos \setminus {\theta}_{1} - {r}_{2} \setminus \cos \setminus {\theta}_{2}\right)}^{2} + {\left({r}_{1} \setminus \sin \setminus {\theta}_{1} - {r}_{2} \setminus \sin \setminus {\theta}_{2}\right)}^{2}}$

$= \setminus \sqrt{{r}_{1}^{2} \setminus {\cos}^{2} \setminus {\theta}_{1} + {r}_{2}^{2} \setminus {\cos}^{2} \setminus {\theta}_{2} - 2 {r}_{1} {r}_{2} \setminus \cos \setminus {\theta}_{1} \setminus \cos \setminus {\theta}_{2} + {r}_{1}^{2} \setminus {\sin}^{2} \setminus {\theta}_{1} + {r}_{2}^{2} \setminus {\sin}^{2} \setminus {\theta}_{2} - 2 {r}_{1} {r}_{2} \setminus \sin \setminus {\theta}_{1} \setminus \sin \setminus {\theta}_{2}}$

$= \setminus \sqrt{{r}_{1}^{2} \left(\setminus {\cos}^{2} \setminus {\theta}_{1} + \setminus {\sin}^{2} \setminus {\theta}_{1}\right) + {r}_{2}^{2} \left(\setminus {\cos}^{2} \setminus {\theta}_{2} + \setminus {\sin}^{2} \setminus {\theta}_{2}\right) - 2 {r}_{1} {r}_{2} \left(\setminus \cos \setminus {\theta}_{1} \setminus \cos \setminus {\theta}_{2} + \setminus \sin \setminus {\theta}_{1} \setminus \sin \setminus {\theta}_{2}\right)}$

$= \setminus \sqrt{{r}_{1}^{2} + {r}_{2}^{2} - 2 {r}_{1} {r}_{2} \setminus \cos \left(\setminus {\theta}_{1} - \setminus {\theta}_{2}\right)}$

Hence, the distance $d$ between the given points $\left({r}_{1} , \setminus {\theta}_{1}\right) \setminus \equiv \left(6 , \frac{19 \setminus \pi}{12}\right)$ & $\left({r}_{2} , \setminus {\theta}_{2}\right) \setminus \equiv \left(3 , \frac{15 \setminus \pi}{8}\right)$ is given by substituting the corresponding values in above general formula as follows

$d = \setminus \sqrt{{6}^{2} + {3}^{2} - 2 \setminus \cdot 6 \setminus \cdot 3 \setminus \cos \left(\frac{19 \setminus \pi}{12} - \frac{15 \setminus \pi}{8}\right)}$

$= \setminus \sqrt{45 - 36 \setminus \cos \left(\frac{7 \setminus \pi}{24}\right)}$

$= 4.804642396$