How do you evaluate # e^( (3 pi)/2 i) - e^( ( pi)/12 i)# using trigonometric functions?

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Answer 1 · 2016-06-10T05:22:32

#e^((3pi)/2i)-e^(pi/12i)=-(1+cos(pi/12))+isin(pi/12)#

We can write #e^(itheta)=costheta+isintheta#

Hence #e^((3pi)/2i)-e^(pi/12i)#

= #[cos((3pi)/2)+isin((3pi)/2)]-[cos(pi/12)+isin(pi/12)]#

But #cos((3pi)/2)=0# and #sin((3pi)/2)=-1#

Hence #e^((3pi)/2i)-e^(pi/12i)#

= #[0-1]-[cos(pi/12)+isin(pi/12)]#

= #-(1+cos(pi/12))+isin(pi/12)#