# How do you find the distance between the points with the given polar coordinates P_1(-2.5,pi/8) and P_2(-1.75, -(2pi)/5)?

May 11, 2018

We get the awesome exact answer

$\sqrt{\frac{149}{16} + \frac{35}{8} \sqrt{\frac{1}{2} \left(4 - \sqrt{2 \left(4 + \sqrt{2 \left(5 + \sqrt{5}\right)}\right)}\right)}}$

#### Explanation:

There's no magic; we convert to rectangular and do the square root of the sum of squared differences. Let's do it generally and rename the points $P = {P}_{1} \left(p , a\right)$ and $Q = {P}_{2} \left(q , b\right)$.

$| P Q | = \setminus \sqrt{{\left({P}_{x} - {Q}_{x}\right)}^{2} + {\left({P}_{y} - {Q}_{y}\right)}^{2}}$

$= \setminus \sqrt{{\left(p \cos a - q \cos b\right)}^{2} + {\left(p \sin a - q \sin b\right)}^{2}}$

That's it.

For the current problem we substitute

$p = - \frac{5}{2} \quad \quad a = \frac{\pi}{8} \quad \quad q = - \frac{7}{4} \quad \quad b = - \frac{2 \pi}{5}$.

The negative magnitudes don't need special handling.

$| P Q | = \setminus \sqrt{{\left(\left(- \frac{5}{2}\right) \cos \left(\frac{\pi}{8}\right) - \left(- \frac{7}{4}\right) \cos \left(\frac{- 2 \pi}{5}\right)\right)}^{2} + {\left(\left(- \frac{5}{2}\right) \sin \left(\frac{\pi}{8}\right) - \left(- \frac{7}{4}\right) \sin \left(\frac{- 2 \pi}{5}\right)\right)}^{2}}$

I'm reasonably sure the your teacher will accept some calculator approximation. But here on Socratic you deserve the exact radical form.

$\cos \left(\frac{\pi}{8}\right) = \cos \left({22.5}^{\circ}\right) = \cos \left({45}^{\circ} / 2\right) = \sqrt{\frac{1}{2} \left(1 + \cos {45}^{\circ}\right)} = \sqrt{\frac{1}{2} \left(1 + \frac{\sqrt{2}}{2}\right)} = \frac{1}{2} \sqrt{2 + \sqrt{2}}$

$\sin \left(\frac{\pi}{8}\right) = \sqrt{\frac{1}{2} \left(1 - \cos {45}^{\circ}\right)} = \frac{1}{2} \sqrt{2 - \sqrt{2}}$

I just did this other answer and got

$\cos \left(- \frac{2 \pi}{5}\right) = \cos {72}^{\circ} = \frac{1}{4} \left(\sqrt{5} - 1\right)$

so I won't repeat that here. There's no shortcut to the sine, we simply get

$\sin \left(- \frac{2 \pi}{5}\right) = - \setminus \sqrt{1 - {\cos}^{2} \left(- \frac{2 \pi}{5}\right)} = - \sqrt{1 - {\left(\frac{1}{4} \left(\sqrt{5} - 1\right)\right)}^{2}} = - \frac{1}{4} \sqrt{10 + 2 \sqrt{5}}$

The minus sign comes because we're in the fourth quadrant.

Putting it all together,

$| P Q | = \setminus \sqrt{{\left(\left(- \frac{5}{2}\right) \left(\frac{1}{2} \sqrt{2 + \sqrt{2}}\right) - \left(- \frac{7}{4}\right) \left(\frac{1}{4} \left(\sqrt{5} - 1\right)\right)\right)}^{2} + {\left(\left(- \frac{5}{2}\right) \left(\frac{1}{2} \sqrt{2 - \sqrt{2}}\right) - \left(- \frac{7}{4}\right) \left(- \frac{1}{4} \sqrt{10 + 2 \sqrt{5}}\right)\right)}^{2}}$

$| P Q | = \sqrt{\frac{149}{16} + \frac{35}{8} \sqrt{\frac{1}{2} \left(4 - \sqrt{2 \left(4 + \sqrt{2 \left(5 + \sqrt{5}\right)}\right)}\right)}}$

$| P Q | \approx 3.1621222$

Check:

Let's pop the original expression into Alpha . Approximation looks right!