How do you convert # (12, 7pi/4)# to rectangular form?

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Answer 1 · 2016-02-20T06:31:59

#x=12 cos ((7pi)/4)=12*(1/sqrt2)=6sqrt2#

#y=12 sin ((7pi)/4)=12*(-1/sqrt2)=-6sqrt2#

Rectangular Form #(6sqrt2, -6sqrt2)#

Use #x=r cos theta#
Use #y=r sin theta#

#x=12 cos ((7pi)/4)=12*(1/sqrt2)=6sqrt2#

#y=12 sin ((7pi)/4)=12*(-1/sqrt2)=-6sqrt2#

Rectangular Form #(6sqrt2, -6sqrt2)#

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