# What is the distance between (2 ,(3 pi)/4 ) and (2 , (13 pi )/8 )?

Sep 20, 2016

$4 \cos \left(\frac{\pi}{16}\right) \approx 4 \left(0.9808\right) = 3.9232$.

#### Explanation:

The Distance btwn. pts. $P \left({r}_{1} , {\theta}_{1}\right) \mathmr{and} Q \left({r}_{2} , {\theta}_{2}\right)$, i.e.,

$P Q$, is given by,

$P {Q}^{2} = {r}_{1}^{2} + {r}_{2}^{2} - 2 {r}_{1} \cdot {r}_{2} \cdot \cos \left({\theta}_{1} - {\theta}_{2}\right)$.

In our case, if $d$ is the reqd. dist., then,

${d}^{2} = {2}^{2} + {2}^{2} - 2 \cdot 2 \cdot 2 \cos \left(13 \frac{\pi}{8} - 3 \frac{\pi}{4}\right)$

$= 8 - 8 \cos \left(7 \frac{\pi}{8}\right)$

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

$= 8 + 8 \cos \left(\frac{\pi}{8}\right)$

$= 8 \left(1 + \cos \left(\frac{\pi}{8}\right)\right)$

$= 8 \cdot 2 {\cos}^{2} \left(\frac{1}{2} \cdot \frac{\pi}{8}\right) = 16 {\cos}^{2} \left(\frac{\pi}{16}\right)$.

$\therefore d = 4 \cos \left(\frac{\pi}{16}\right)$.

Numerically, $d \approx 4 \left(0.9808\right) = 3.9232$.