What is the Cartesian form of #( -2 , (13pi)/6 ) #?

1 Answer
Jan 19, 2017

I got:

#(sqrt3,1)#

assuming #r > 0#. If #r < 0#, then we would necessarily have #r = -sqrt(x^2 + y^2)# to keep the equations below self-consistent, and then we would have #x = -sqrt3# and #y = -1#.


Recall:

#x = rcostheta#
#y = rsintheta#
#r = sqrt(x^2 + y^2)#

The coordinates you are given are in polar, and are the #r# and #theta# coordinates. Thus, you need to use the expression for #r# to calculate #x# or #y#, and the expression for #theta# to get the other coordinate.

However... #r# is only positive or #0#. You cannot have a negative radius, as it is physically impossible. So, there's a typo in the problem. It should be #(2,(13pi)/6)#.

#2 = sqrt(x^2 + y^2) => 4 = x^2 + y^2#

#y = rsin(theta) = 2sin((13pi)/6)#

#=> y^2 = 4sin^2((13pi)/6) = 4sin^2(390^@) = 4sin^2(30^@) = 1#

#=> color(green)(y = 1)#

(to keep consistent with the equations above, #y > 0#, since #costheta > 0# and #r > 0#.)

Therefore, we can solve for #x#:

#=> x^2 = 4 - y^2 = 4 - 1 = 3#

#=> color(green)(x = sqrt3)#

(to keep consistent with the equations above, #xy > 0#, since #costheta > 0# and #r > 0#.)

So, the cartesian coordinates are:

#color(blue)((x,y) = (sqrt3,1))#

To make sure it worked...

#r = sqrt(3 + 1) = 2# #color(blue)(sqrt"")#

#theta = arccos(x/r) = arcsin(y/r)#

#= arccos(sqrt3/2)#

As #costheta = sqrt3/2# when #theta = 30^@#, #arccos(sqrt3/2) = theta = 30^@#, which is coterminal with #390^@#. #color(blue)(sqrt"")#