How do you convert the polar coordinate (6, 2pi/3) into cartesian coordinates?

1 Answer
Aug 4, 2015

#(x, y) = (r cos theta, r sin theta) = (6cos((2pi)/3), 6sin((2pi)/3))#

#= (6*-1/2, 6*sqrt(3)/2) = (-3,3sqrt(3))#

Explanation:

Given radius #r# and angle #theta#, the cartesian coordinates #x# and #y# are given by the formulae:

#x = r cos theta#
#y = r sin theta#

Conversely, given #x# and #y#, the radius #r# and angle #theta# are determined by the formulae:

#r = sqrt(x^2+y^2)#
#theta = "atan2"(y, x)#

where #"atan2"(y, x)# is defined as follows:

#"atan2"(y, x) = { (arctan(y/x), " if x > 0"), (arctan(y/x) + pi, " if x < 0 and y >= 0"), (arctan(y/x) - pi, " if x < 0 and y < 0"), (pi/2, " if x = 0 and y > 0"), (-pi/2, " if x = 0 and y < 0"), ("undefined", " if x = 0 and y = 0") :}#