# What is the distance between the following polar coordinates?:  (2,(3pi)/4), (1,(15pi)/8)

##### 3 Answers
Jul 18, 2018

$D = \sqrt{5 + 4 \cos \left(\frac{\pi}{8}\right)} \approx 2.9488$

#### Explanation:

We know that ,

$\text{Distance between Polar Co-ordinates:} A \left({r}_{1} , {\theta}_{1}\right) \mathmr{and} B \left({r}_{2} , {\theta}_{2}\right)$ is

color(red)(D=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_1-theta_2))...to(I)

We have , ${P}_{1} \left(2 , \frac{3 \pi}{4}\right) \mathmr{and} {P}_{2} \left(1 , \frac{15 \pi}{8}\right)$.

So , ${r}_{1} = 2 , {r}_{2} = 1 , {\theta}_{1} = \frac{3 \pi}{4} \mathmr{and} {\theta}_{2} = \frac{15 \pi}{8}$

$\implies {\theta}_{1} - {\theta}_{2} = \frac{3 \pi}{4} - \frac{15 \pi}{8} = \frac{6 \pi - 15 \pi}{8} = \frac{- 9 \pi}{8}$

$\implies \cos \left({\theta}_{1} - {\theta}_{2}\right) = \cos \left(\frac{- 9 \pi}{8}\right)$
$\implies \cos \left({\theta}_{1} - {\theta}_{2}\right) = \cos \left(\frac{9 \pi}{8}\right) \to \left[\because \cos \left(- \theta\right) = \cos \theta\right]$
$\implies \cos \left({\theta}_{1} - {\theta}_{2}\right) = \cos \left(\pi + \frac{\pi}{8}\right) = - \cos \left(\frac{\pi}{8}\right)$

"Using : " color(red)((I) we get

D=sqrt(2^2+1^2-2(2)(1)(-cos(pi/8))

$\implies D = \sqrt{4 + 1 + 4 \cdot \cos \left(\frac{\pi}{8}\right)}$

$\implies D = \sqrt{5 + 4 \cos \left(\frac{\pi}{8}\right)}$

$\implies D \approx 2.9488$

Jul 18, 2018

$\sqrt{2 + \sqrt{2 - \sqrt{2}}}$

#### Explanation:

It is nontrivial to determine polar distances (since drawing a line in polar coordinates isn't very easy), so we should convert to Cartesian coordinates.

The first point is moderately easy:
$x = r \cos \theta = 2 \cdot - \frac{\sqrt{2}}{2} = - \sqrt{2}$
$y = r \sin \theta = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$

The second point is a little harder, but we only need to use half angle identities. Let $\theta = \frac{15 \pi}{4} \equiv - \frac{\pi}{4}$. Therefore, $\cos \left(\theta\right) = \frac{\sqrt{2}}{2}$ and

$\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} = \pm \frac{\sqrt{2 + \sqrt{2}}}{2}$
$\sin \left(\frac{\theta}{2}\right) = \pm \frac{\sqrt{1 - \cos \theta}}{2} = \pm \frac{\sqrt{2 - \sqrt{2}}}{2}$

Since $\frac{15 \pi}{8}$ is in the 4th quadrant, its cosine is positive and its sine is negative. Therefore,

$x = r \cos \left(\frac{15 \pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$
$y = r \sin \left(\frac{15 \pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$

This allows us to use Pythagorean theorem in order to find the distance between the two points:

$\Delta x = \frac{\sqrt{2 + \sqrt{2}}}{2} + \sqrt{2}$
$\Delta y = \frac{\sqrt{2 - \sqrt{2}}}{2} - \sqrt{2}$

Pulling out a factor of $\sqrt{2}$ from each, then squaring and adding them,
${\mathrm{ds}}^{2} = 2 \left[1 + \frac{\sqrt{2}}{2} + 1 + 1 - \frac{\sqrt{2}}{2} + 1 + 2 \sqrt{1 + \frac{\sqrt{2}}{2}} - 2 \sqrt{1 - \frac{\sqrt{2}}{2}}\right]$
$= 2 \left[4 + 2 \left(\sqrt{1 + \frac{\sqrt{2}}{2}} - \sqrt{1 - \frac{\sqrt{2}}{2}}\right)\right]$
Using the cosine sum formula with $\frac{\pi}{8}$ and $\frac{\pi}{4}$, we get
$= 8 \left(1 + \cos \left(\frac{3 \pi}{8}\right)\right) = 4 {\cos}^{2} \left(\frac{3 \pi}{16}\right)$

Therefore the total distance is
$D = 2 \cos \left(\frac{3 \pi}{16}\right) = \sqrt{2 + \sqrt{2 - \sqrt{2}}} \approx 1.66$

$2.949 \setminus \setminus \textrm{u n i t}$

#### Explanation:

The distance between the points $\left({r}_{1} , \setminus {\theta}_{1}\right) \setminus \equiv \left(2 , \frac{3 \setminus \pi}{4}\right)$ & $\left({r}_{2} , \setminus {\theta}_{2}\right) \setminus \equiv \left(1 , \frac{15 \setminus \pi}{8}\right)$ is given by the formula as follows

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

$= \setminus \sqrt{{2}^{2} + {1}^{2} - 2 \left(2\right) \left(1\right) \setminus \cos \left(\frac{3 \setminus \pi}{4} - \frac{15 \setminus \pi}{8}\right)}$

$= \setminus \sqrt{5 - 4 \setminus \cos \left(\frac{9 \setminus \pi}{8}\right)}$

$= \setminus \sqrt{5 + 4 \setminus \cos \left(\frac{\setminus \pi}{8}\right)}$

$= \setminus \sqrt{\setminus \frac{7 \setminus \sqrt{2} + 2}{\setminus \sqrt{2}}}$

$= 2.949 \setminus \setminus \textrm{u n i t}$