What is the distance between the following polar coordinates?:  (6,(7pi)/12), (2,(-3pi)/8)

Mar 15, 2016

d = 2sqrt(10 - 3sqrt( 2 + sqrt ( 2 - sqrt 3 ) )

Explanation:

$d = \sqrt{{r}_{1}^{2} + {r}_{2}^{2} - 2 {r}_{1} {r}_{2} \cos \left({\phi}_{2} - {\phi}_{1}\right)}$

d = sqrt(6^2+2^2 - 2*6*2*cos(-(3pi)/8-(7pi)/12)

d = sqrt(40 - 24*cos((-29pi)/24)

d = 2sqrt(10 - 6cos((29pi)/24)

$\cos \left(\frac{29 \pi}{6}\right) = \cos \left(4 \pi + \frac{5 \pi}{6}\right) = \cos \left(\frac{5 \pi}{6}\right) = - \frac{\sqrt{3}}{2}$

$\cos \left(\frac{29 \pi}{12}\right) = \cos \left(\frac{\frac{29 \pi}{6}}{2}\right) = \sqrt{\frac{1 + \cos \left(\frac{29 \pi}{6}\right)}{2}}$

$\cos \left(\frac{29 \pi}{12}\right) = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$

$\cos \left(\frac{29 \pi}{24}\right) = \cos \left(\frac{\frac{29 \pi}{12}}{2}\right) = \sqrt{\frac{1 + \cos \left(\frac{29 \pi}{12}\right)}{2}}$

$\cos \left(\frac{29 \pi}{24}\right) = \sqrt{\frac{1 + \frac{\sqrt{\left(2 - \sqrt{3}\right)}}{2}}{2}}$

$\cos \left(\frac{29 \pi}{24}\right) = \frac{\sqrt{2 + \sqrt{2 - \sqrt{3}}}}{2}$

$d = 2 \sqrt{10 - 6 \cdot \frac{\sqrt{2 + \sqrt{2 - \sqrt{3}}}}{2}}$

d = 2sqrt(10 - 3sqrt( 2 + sqrt ( 2 - sqrt 3 ) )