How do you evaluate # e^( ( 15 pi)/8 i) - e^( ( 11 pi)/6 i)# using trigonometric functions?
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#e^((15pi)/8i)-e^((11pi)/6i)=-1.7899+0.8827i#
As #e^(itheta)=costheta+isintheta#, we have
#e^((15pi)/8i)=cos((15pi)/8)+isin((15pi)/8)#
= #cos(pi-pi/8)+isin(pi-pi/8)#
= #-cos(pi/8)+isin(pi/8)=-0.9239+0.3827i#
#e^((11pi)/6i)=cos((11pi)/6)+isin((11pi)/6)#
= #cos(2pi-pi/6)+isin(2pi-pi/6)#
= #cos(pi/6)-isin(pi/6)#
= #sqrt3/2-i*1/2=0.8660-0.5i#
Hence #e^((15pi)/8i)-e^((11pi)/6i)=(-0.9239+0.3827i)-(0.8660-0.5i)#
= #(-0.9239-0.8660)+i(0.3827+0.5)#
= #-1.7899+0.8827i#
