Question

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)`?

Answer 1

We get the awesome exact answer

`sqrt(149/16 + 35/8 sqrt(1/2 (4 - sqrt(2 (4 + sqrt(2 (5 + sqrt(5))))))))`

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(p, a)` and `Q=P_2(q,b)`.

`| PQ | = \sqrt{ (P_x - Q_x)^2 + (P_y - Q_y)^2}`

`= \sqrt{ ( p cos a - q cos b)^2 + ( p sin a - q sin b)^2 }`

That's it.

For the current problem we substitute

`p=-5/2 quad quad a=pi/8 quad quad q =-7/4 quad quad b=-{2pi}/5`.

The negative magnitudes don't need special handling.

`| PQ | = \sqrt{ ( (-5/2) cos (pi/8) - (-7/4) cos ({-2pi}/5))^2 + ( (-5/2) sin (pi/8) - (-7/4) sin ({-2pi}/5))^2 }`

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

`cos(pi/8) = cos(22.5^circ)= cos(45^circ / 2) = sqrt{ 1/2 (1 + cos 45^circ) } = sqrt{ 1/2 (1 + sqrt{2}/2) } =1/ 2 sqrt(2 + sqrt(2))`

`sin(pi/8) = sqrt{ 1/2 (1 - cos 45^circ) } =1/ 2 sqrt(2 - sqrt(2))`

I just did this other answer and got

`cos(-{2pi}/5)=cos 72^circ = 1/4 (sqrt{5} - 1 )`

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

`sin(-{2pi}/5)=-\sqrt{1-cos^2(-{2pi}/5)} = -sqrt{ 1- ( 1/4 (sqrt{5} - 1 ))^2 } = -1/4 sqrt(10 + 2 sqrt(5) )`

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

Putting it all together,

`| PQ | = \sqrt{ ( (-5/2) ( 1/ 2 sqrt(2 + sqrt(2) ) ) - (-7/4) (1/4 (sqrt{5} - 1 ) ) )^2 + ( (-5/2) (1/ 2 sqrt(2 - sqrt(2))) - (-7/4) (-1/4 sqrt(10 + 2 sqrt(5) )) )^2 }`

`|PQ| = sqrt(149/16 + 35/8 sqrt(1/2 (4 - sqrt(2 (4 + sqrt(2 (5 + sqrt(5))))))))`

`|PQ| approx 3.1621222`

Check:

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