# What is the distance between (2 ,(7 pi)/6 ) and (3 , (- pi )/8 )?

Jan 3, 2016

$1.0149$

#### Explanation:

The distance formula for polar coordinates is

d=sqrt(r_1^2+r_2^2-2r_1r_2Cos(theta_1-theta_2)
Where $d$ is the distance between the two points, ${r}_{1}$, and ${\theta}_{1}$ are the polar coordinates of one point and ${r}_{2}$ and ${\theta}_{2}$ are the polar coordinates of another point.
Let $\left({r}_{1} , {\theta}_{1}\right)$ represent $\left(2 , \frac{7 \pi}{6}\right)$ and $\left({r}_{2} , {\theta}_{2}\right)$ represent $\left(3 , - \frac{\pi}{8}\right)$.
implies d=sqrt(2^2+3^2-2*2*3Cos((7pi)/6-(-pi/8))
implies d=sqrt(4+9-12Cos((7pi)/6+pi/8)
$\implies d = \sqrt{13 - 12 \cos \left(\frac{28 \pi + 3 \pi}{24}\right)} = \sqrt{13 - 12 \cos \left(\frac{31 \pi}{24}\right)} = \sqrt{13 - 12 \cos \left(4.0558\right)} = \sqrt{13 - 12 \cdot 0.9975} = \sqrt{13 - 12 \cdot 0.9975} = \sqrt{13 - 11.97} = \sqrt{1.03} = 1.0149$ units
$\implies d = 1.0149$ units (approx)
Hence the distance between the given points is $1.0149$.