How do you evaluate # e^( ( 13 pi)/8 i) - e^( ( 7 pi)/12 i)# using trigonometric functions?

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Answer 1 · 2016-03-28T06:40:42

#e^((13pi)/8i)-e^((7pi)/12i)=(0.6415+0.042i)#

As #e^(itheta)=costheta+isintheta#, we have

#e^((13pi)/8i)=cos((13pi)/8)+isin((13pi)/8)# and

#e^((7pi)/12i)=cos((7pi)/12)+isin((7pi)/12)#

Hence, #e^((13pi)/8i)-e^((7pi)/12i)#

= #(cos((13pi)/8)+isin((13pi)/8))-(cos((7pi)/12)+isin((7pi)/12))#

As #cos(13pi/8)=0.3827#, #sin(13pi/8)=-0.9239#,

#cos(7pi/12)=-0.2588# and #sin((7pi)/12)=-0.9659#

#e^((13pi)/8i)-e^((7pi)/12i)#

= #(0.3827+i(-0.9239))-((-0.2588)+i(-0.9659)#

= #(0.6415+0.042i)#