# What is the distance between the following polar coordinates?:  (6,(7pi)/12), (3,(-5pi)/8)

Nov 10, 2016

graph{((x+3/2sqrt(2+sqrt(2)))^2+(y-3/2sqrt(2-sqrt(2)))^2-.01)((x+3sqrt(2+sqrt(3)))^2+(y+3sqrt(2-sqrt(3)))^2-.01)=0 [-10.71, 5.1, -2.67, 5.23]}

$d = 3 \sqrt{5 - \sqrt{8 + 2 \left(\sqrt{6} - \sqrt{2}\right)}}$

#### Explanation:

$\cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{2 + \sqrt{3}}}{2}$
$\sin \left(\frac{\pi}{12}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$
$\cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$
$\sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$

so

P_1=(-3sqrt(2+sqrt(3));-3sqrt(2-sqrt(3)))
P_2=(-3/2sqrt(2+sqrt(2));3/2sqrt(2-sqrt(2)))

$\overline{{P}_{1} {P}_{2}} = \sqrt{{\left(- 3 \sqrt{2 + \sqrt{3}} + \frac{3}{2} \sqrt{2 + \sqrt{2}}\right)}^{2} + {\left(3 \sqrt{2 - \sqrt{3}} + \frac{3}{2} \sqrt{2 - \sqrt{2}}\right)}^{2}} =$
$= \sqrt{18 + \cancel{9 \sqrt{3}} + \frac{9}{2} + \cancel{\frac{9}{4} \sqrt{2}} - 9 \sqrt{\left(2 + \sqrt{3}\right) \left(2 + \sqrt{2}\right)} + 18 - \cancel{9 \sqrt{3}} + \frac{9}{2} - \cancel{\frac{9}{4} \sqrt{2}} + 9 \sqrt{\left(2 - \sqrt{2}\right) \left(2 - \sqrt{3}\right)}} =$
$= \sqrt{45 - 9 \left(\sqrt{4 + \sqrt{6} + 2 \left(\sqrt{2} + \sqrt{3}\right)} - \sqrt{4 + \sqrt{6} - 2 \left(\sqrt{2} + \sqrt{3}\right)}\right)}$
$= 3 \sqrt{5 - \left(\sqrt{4 + \sqrt{6} + 2 \sqrt{5 + 2 \sqrt{6}}} - \sqrt{4 + \sqrt{6} - 2 \sqrt{5 + 2 \sqrt{6}}}\right)}$

Now put
$a = 8 + 2 \sqrt{6}$
and
$b = 8$
then
${a}^{2} - b = 80 + 32 \sqrt{6}$
so using
$\sqrt{a - \sqrt{b}} = \sqrt{\frac{a + \sqrt{{a}^{2} - b}}{2}} - \sqrt{\frac{a + \sqrt{{a}^{2} - b}}{2}}$
$\sqrt{8 + 2 \sqrt{6} - \sqrt{8}} =$
$= \sqrt{\frac{8 + 2 \sqrt{6} + 4 \sqrt{5 + 2 \sqrt{6}}}{2}} - \sqrt{\frac{8 + 2 \sqrt{6} - 4 \sqrt{5 + 2 \sqrt{6}}}{2}}$
$= \sqrt{4 + \sqrt{6} + 2 \sqrt{5 + 2 \sqrt{6}}} - \sqrt{4 + \sqrt{6} - 2 \sqrt{5 + 2 \sqrt{6}}}$

So

$\overline{{P}_{1} {P}_{2}} = 3 \sqrt{5 - \sqrt{8 + 2 \sqrt{6} - \sqrt{8}}} = 3 \sqrt{5 - \sqrt{8 + 2 \left(\sqrt{6} - \sqrt{2}\right)}}$