# What is the distance between the following polar coordinates?:  (-4,(3pi)/4), (5,(3pi)/8)

Jun 16, 2018

$l = \sqrt{41 - 20 \sqrt{2}}$

#### Explanation:

Let's solve a general case; say we have two points of Polar coordinates $A \left({r}_{1} , {\varphi}_{1}\right)$ and $B \left({r}_{2} , {\varphi}_{2}\right)$. Let us visualise the situation:

Let $l$ be the distance between $A$ and $B$:

$l = A B$

We might convert $A$ and $B$ to Cartesian coordinates, for which we have a formula, but instead let's keep everything in Polar.

In the triangle $\Delta A O B$, we know the lenght of the sides $O A$, $O B$ and the angle between the; $\angle A O B$.

As such, we can apply the cosine law:

$A {B}^{2} = O {A}^{2} + O {B}^{2} - 2 O A \cdot O B \cos \left(\angle A O B\right)$

Substituting everything into Polar coordinates:

${l}^{2} = {r}_{1}^{2} + {r}_{2}^{2} - 2 {r}_{1} {r}_{2} \cos \left({\varphi}_{1} - {\varphi}_{2}\right)$

Finally, we can solve our particular case:

$A \left(- 4 , \frac{3 \pi}{4}\right) \mathmr{and} B \left(5 , \frac{3 \pi}{8}\right)$

$\therefore {l}^{2} = {\left(- 4\right)}^{2} + {5}^{2} - 2 \left(- 4\right) \cdot 5 \cos \left(\frac{3 \pi}{4} - \frac{3 \pi}{8}\right)$

$\therefore {l}^{2} = 16 + 25 + 40 \cos \left(\frac{3 \pi}{4}\right)$

Notice how $\frac{3 \pi}{4} = \pi - \frac{\pi}{4}$, hence

$\cos \left(\frac{3 \pi}{4}\right) = - \cos \left(\frac{\pi}{4}\right) = - \frac{\sqrt{2}}{2}$

$\therefore {l}^{2} = 41 - 40 \cdot \frac{\sqrt{2}}{2} = 41 - 20 \sqrt{2}$

$\therefore l = \sqrt{41 - 20 \sqrt{2}}$