What is the Cartesian form of #( 4 , ( 11pi)/4 ) #?

1 Answer
Mar 22, 2016

#(-2sqrt2 , 2sqrt2 )#

Explanation:

Using the formulae that links Polar to Cartesian coordinates.

#• x = rcostheta #

#• y = rsintheta #

here r = 4 and # theta = (11pi)/4#
# "---------------------------------------------------"#
#rArr x = 4cos((11pi)/4) = 4cos((11pi)/4 - 2pi) = cos((3pi)/4)#

and # cos((3pi)/4) = -cos(pi/4) = - 1/sqrt2 #
thus # x = 4xx(-1)/sqrt2 = - 2sqrt2 #
# "-------------------------------------------------"#

# rArr y = 4sin((11pi)/4) = 4sin(pi/4) = 4xx1/sqrt2 = 2sqrt2#