What is the Cartesian form of #(-2,(-11pi) /16)#?

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Answer 1 · 2018-06-11T05:30:02

Cartesian form #(-2*cos((-11pi)/16), -2*sin((-11pi)/16))=(1.111, 1.663)#

The given #(r, theta)=(-2, -(11pi)/16)#

using

#x=r*cos theta=-2*cos((-11pi)/16)=1.111#

#y=r*sin theta=-2*sin((-11pi)/16)=1.663#

and our Cartesian form #(x, y)=(1.111, 1.663)#

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I hope the explanation is useful....God bless..