Question

What is the distance between the following polar coordinates?: `(2,(3pi)/4), (1,(15pi)/8)`

Answer 1

`D=sqrt(5+4cos(pi/8))~~2.9488`

Explanation:

We know that ,

`"Distance between Polar Co-ordinates:"A(r_1,theta_1)and B(r_2,theta_2)` is

`color(red)(D=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_1-theta_2))...to(I)`

We have , `P_1(2,(3pi)/4) and P_2(1,(15pi)/8)`.

So , `r_1=2 , r_2=1 , theta_1=(3pi)/4 and theta_2=(15pi)/8`

`=>theta_1-theta_2=(3pi)/4-(15pi)/8=(6pi-15pi)/8=(-9pi)/8`

`=>cos(theta_1-theta_2)=cos((-9pi)/8)`
`=>cos(theta_1-theta_2)=cos((9pi)/8)to[becausecos(-theta)=costheta]`
`=>cos(theta_1-theta_2)=cos(pi+pi/8)=-cos(pi/8)`

`"Using : " color(red)((I)` we get

`D=sqrt(2^2+1^2-2(2)(1)(-cos(pi/8))`

`=>D=sqrt(4+1+4*cos(pi/8))`

`=>D=sqrt(5+4cos(pi/8))`

`=>D~~2.9488`

Answer 2

`sqrt(2+sqrt(2-sqrt(2)))`

Explanation:

It is nontrivial to determine polar distances (since drawing a line in polar coordinates isn't very easy), so we should convert to Cartesian coordinates.

The first point is moderately easy:
`x = rcostheta = 2 * -sqrt(2)/2 = -sqrt(2)`
`y = rsintheta = 2 * sqrt(2)/2 = sqrt(2)`

The second point is a little harder, but we only need to use half angle identities. Let `theta = (15pi)/4 equiv -(pi)/4`. Therefore, `cos(theta) = sqrt(2)/2` and

`cos(theta/2) = pm sqrt((1+costheta)/2) = pm sqrt(2+sqrt2)/2`
`sin(theta/2) = pm sqrt(1-costheta)/2 = pm sqrt(2-sqrt2)/2`

Since `(15pi)/8` is in the 4th quadrant, its cosine is positive and its sine is negative. Therefore,

`x = r cos((15pi)/8) = sqrt(2+sqrt2)/2`
`y = r sin((15pi)/8) = sqrt(2-sqrt2)/2`

This allows us to use Pythagorean theorem in order to find the distance between the two points:

`Delta x = sqrt(2+sqrt2)/2 + sqrt2`
`Delta y = sqrt(2-sqrt2)/2 - sqrt2`

Pulling out a factor of `sqrt(2)` from each, then squaring and adding them,
`ds^2 = 2[1 + sqrt2/2 + 1 + 1 - sqrt2/2 + 1 + 2sqrt(1+sqrt2/2) - 2 sqrt(1 - sqrt2/2)]`
`= 2[4 + 2(sqrt(1+sqrt2/2) - sqrt(1-sqrt2/2))]`
Using the cosine sum formula with `pi/8` and `pi/4`, we get
`= 8 (1 + cos ((3pi)/8) ) = 4 cos^2((3pi)/16)`

Therefore the total distance is
`D = 2 cos((3pi)/16) = sqrt(2 + sqrt(2 - sqrt(2))) approx 1.66`

Answer 3

`2.949\ \text{unit}`

Explanation:

The distance between the points `(r_1, \theta_1)\equiv(2, {3\pi}/4)` & `(r_2, \theta_2)\equiv(1, {15\pi}/8)` is given by the formula as follows

`\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_1-\theta_2)}`

`=\sqrt{2^2+1^2-2(2)(1)\cos({3\pi}/4-{15\pi}/8)}`

`=\sqrt{5-4\cos({9\pi}/8)}`

`=\sqrt{5+4\cos({\pi}/8)}`

`=\sqrt{\frac{7\sqrt2+2}{\sqrt2}}`

`=2.949\ \text{unit}`