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

1 Answer
May 26, 2017

In Cartesian coordinates, #(-4,(11pi)/4)# = #(2sqrt2, -2sqrt2)#

Explanation:

If you consider that #x=rcosTheta# and #y=rsinTheta#, you can plug in those values from that polar coordinate to find #(x,y)#.

On the unit circle, #theta = (11pi)/4# is equal to #(3pi)/4#.

When you plug it in, #x=-4cos((3pi)/4)# and #y=-4sin((3pi)/4)#.

More simplified, #x=-4*-(sqrt2)/2# and #y=-4*(sqrt2)/2#.

Once everything is finally simplified, you will get #x=2sqrt2# and #y = -2sqrt2#