What is the distance between the following polar coordinates?:  (3,(12pi)/8), (9,(5pi)/8)

Mar 31, 2018

$d \approx 9.97$

Explanation:

We can convert each to Cartesian:

$\left(3 , \frac{3 \pi}{2}\right) \rightarrow 3 {e}^{i \frac{3 \pi}{2}} = 3 \left(- i\right) = - 3 i$
$\left(9 , \frac{5 \pi}{8}\right) \rightarrow 9 {e}^{i \frac{5 \pi}{8}} = - \frac{9 \sqrt{2 - \sqrt{2}}}{2} + i \frac{9 \sqrt{2 + \sqrt{2}}}{2}$

In order to do the trig functions for $\frac{5 \pi}{8}$, I used the half angle formulae and the angle addition formulae (just to be careful about signs):

$\frac{5 \pi}{8} = \frac{\pi}{2} + \frac{\pi}{8}$

$\cos \left(\frac{\pi}{8}\right) = \cos \left(\frac{1}{2} \left(\frac{\pi}{4}\right)\right) = \sqrt{\frac{1 + \cos \left(\frac{\pi}{4}\right)}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

$\sin \left(\frac{\pi}{8}\right) = \sin \left(\frac{1}{2} \left(\frac{\pi}{4}\right)\right) = \sqrt{\frac{1 - \cos \left(\frac{\pi}{4}\right)}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$

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

Now, we can just apply Pythagorean theorem:

${d}^{2} = \Delta {x}^{2} + \Delta {y}^{2}$
$= \frac{81}{4} \cdot \left(2 - \sqrt{2}\right) + {\left(\frac{9 \sqrt{2 + \sqrt{2}}}{2} + 3\right)}^{2}$
$= \frac{81}{4} \cdot \left(2 - \sqrt{2}\right) + \frac{81}{4} \cdot \left(2 + \sqrt{2}\right) + 9 + 2 \cdot 3 \cdot \frac{9}{2} \cdot \sqrt{2 + \sqrt{2}}$
$= \frac{81}{2} + \frac{18}{2} + 27 \sqrt{2 + \sqrt{2}}$
${d}^{2} = \frac{99}{2} + 27 \sqrt{2 + \sqrt{2}}$

Which means that the distance is actually

$d = \sqrt{\frac{99}{2} + 27 \sqrt{2 + \sqrt{2}}} \approx 9.97$

Apr 1, 2018

Distance=$18$ units.

Explanation:

Polar coordinates are written as $\left[r , \theta\right]$. From Pythagoras, ${x}^{2} + {y}^{2} = {r}^{2}$....and from the same triangle, $x = r \cos \theta$, $y = r \sin \theta$

So $\left[3 , \frac{12 \pi}{8}\right]$= means $x = 3 \cos 12 \frac{\pi}{8}$ and $y = 3 \sin 12 \frac{\pi}{8}$

So coordinates of this point is$\left[0 , 1\right]$