What is the Cartesian form of #(-2,(-161pi) /16)#?

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Answer 1 · 2017-04-16T00:54:35

#(1.962,-.390)#

Use the parametric form to find the #x# and #y# components of this polar coordinate. Remember that:

#x=rcostheta#

#y=rsintheta#

Plug into these formulas:

#x=-2cos(-161pi/16)~~1.962#

#y=-2sin(-161pi/16)~~-.390#

We have our #x# and #y# component. Now write it as a coordinate:

#(1.962,-.390)#