How do you simplify #3(cos((7pi)/3)+isin((7pi)/3))div(cos(pi/2)+isin(pi/2))# and express the result in rectangular form?

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Answer 1 · 2017-02-18T06:13:14

The answer is #=-3/2+i3sqrt3/2#

We need Euler's relation

#e^(itheta)=costheta+isintheta#

#cos(2/3pi)=-1/2#

#sin(2/3pi)=sqrt3/2#

Therefore,

#cos(7/6pi)+isin(7/6pi)=e^(7/6ipi)#

#cos(pi/2)+isin(pi/2)=e^(ipi/2)#

So,

#(3(cos(7/6pi)+isin(7/6pi)))/(cos(pi/2)+isin(pi/2))#

#=3e^(7/6ipi)/e^(ipi/2)#

#=3e^(ipi(7/6-1/2))#

#=3e^(2/3ipi)#

#=3cos(2/3pi)+i3sin(2/3pi)#

#=-3/2+i3sqrt3/2#