How do you convert # (-pi, -3pi/4)# to rectangular form?

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Answer 1 · 2016-11-17T14:25:54

The answer is #=(pi/sqrt2,pi/sqrt2)#

To convert from polar coordinates #(r,theta)# to rectangular coordinates #(x,y)#, we use the following equations

#x=rcos theta#

#y=rsintheta#

Here, #(r,theta)=(-pi,-3pi/4)#

Therefore, #x=-pi*cos((-3pi)/4)=-pi*-sqrt2/2=pi/sqrt2#

#y=-pi*sin ((-3pi)/4)=-pi*-sqrt2/2=pi/sqrt2#

So the rectangular coordinates are #(x,y)=(pi/sqrt2,pi/sqrt2)#