How do you find the rectangular form of #(4, -pi/2)#?

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Answer 1 · 2018-05-25T04:28:06

#(0, -4)#

Polar coordinates are represented as:

#(r, theta)#

Since we're given:

#(4, -pi/2)#

#4# is #r# and #-pi/2# is #theta#

Use the formula:

#(rcostheta, rsintheta)#

The rest is plugging and solving:

#rcostheta#
#4cos(-pi/2)#
#4(0)# #color(blue)(" The "cos " value of "-pi/2" is 0")#
#0#
#x=0#

#rsintheta#
#4sin(-pi/2)#
#4(-1)# #color(blue)(" The "sin " value of "-pi/2" is -1")#
#-4#
#y=-4#

#color(red)((0, -4)#