What is the distance between the following polar coordinates?: # (3,(12pi)/8), (9,(5pi)/8) #

2 Answers
Mar 31, 2018

#d approx 9.97#

Explanation:

We can convert each to Cartesian:

#(3, (3pi)/2) rightarrow 3e^(i(3pi)/2) = 3(-i) = -3i #
#(9, (5pi)/8) rightarrow 9e^(i(5pi)/8) = -(9sqrt(2-sqrt(2)))/2 + i (9sqrt(2+sqrt(2)))/2 #

In order to do the trig functions for #(5pi)/8#, I used the half angle formulae and the angle addition formulae (just to be careful about signs):

#(5pi)/8 = pi/2 + pi/8 #

#cos(pi/8) = cos(1/2 (pi/4)) = sqrt( (1+cos(pi/4)) /2) = sqrt( 2 + sqrt(2))/2 #

#sin(pi/8) = sin(1/2 (pi/4)) = sqrt( (1-cos(pi/4)) /2) = sqrt( 2 - sqrt(2))/2 #

#cos(pi/2 + pi/8) = - sin(pi/8) = -sqrt( 2- sqrt(2))/2 #
#sin(pi/2 + pi/8) = cos(pi/8) = sqrt(2 + sqrt(2))/2 #

Now, we can just apply Pythagorean theorem:

#d^2 = Delta x^2 + Delta y^2 #
#= 81/4 * (2 - sqrt(2)) + ((9sqrt(2+sqrt2))/2 + 3)^2 #
#= 81/4 * (2 - sqrt(2)) + 81/4 * (2+sqrt2) + 9 + 2 * 3 * 9/2 *sqrt(2+sqrt2) #
#= 81/2 + 18/2 + 27sqrt(2+sqrt2) #
#d^2= 99/2 + 27sqrt(2+sqrt2) #

Which means that the distance is actually

#d = sqrt(99/2 + 27sqrt(2+sqrt2)) approx 9.97 #

Apr 1, 2018

Distance=#18# units.

Explanation:

Polar coordinates are written as #[r,theta]#. From Pythagoras, #x^2+y^2=r^2#....and from the same triangle, #x=rcostheta#, #y=rsintheta#

So #[3, [12pi]/[8]]#= means #x=3cos12pi/[8]# and #y=3sin12pi/[8]#

So coordinates of this point is#[0,1]#