# What is the distance between the following polar coordinates?:  (2,(13pi)/4), (7,(-3pi)/8)

Jan 10, 2017

$\sqrt{53 - 28 \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}}} = 6.503 ,$ nearly.

#### Explanation:

Let the points be $P \left(2 , \frac{13}{4} \pi\right) \mathmr{and} Q \left(7 , - \frac{3}{5} \pi\right)$.

The position-vector $\vec{O P}$direction $\theta = \frac{13}{4} \pi = 4 \pi - \frac{3}{4} \pi$.

This is same as is the same as $\theta = - \frac{3}{4} \pi$.

So, as far as the position is concerned, the first point P is (1, -3/4pi).

Now, the $\angle P O Q = \left(- \frac{3}{8} \pi\right) - \left(- \frac{3}{4} \pi\right) = \frac{3}{8} \pi$,

The lengths-relation for the sides of the $\triangle P O Q$ is

$P Q = \sqrt{O {P}^{2} + O {Q}^{2} - 2 \left(O P\right) \left(O P\right) \cos \angle P O Q}$

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

$\cos \left(\frac{3}{8} \pi\right) = \sin \left(\frac{\pi}{2} - \frac{3}{8} \pi\right) = \sin \left(\frac{\pi}{8}\right) = \sqrt{\frac{1 - \cos \left(2 \left(\frac{\pi}{8}\right)\right)}{2}}$

=sqrt(((1-1/sqrt2)/2). And so,

$P Q = \sqrt{53 - 28 \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}}} = 6.503 ,$ nearly.