How do you evaluate # e^( ( pi)/4 i) - e^( ( 3 pi)/8 i)# using trigonometric functions?

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Answer 1 · 2016-03-23T02:47:11

#e^(pi/4 i) -e^((3pi)/8 i) ~~0.324-i0.217#

#e^(itheta)=cos theta + i sin theta#

#theta_1=pi/4, theta_2 = (3pi)/8#

#e^(pi/4 i) =cos (pi/4) +isin(pi/4) #

#e^((3pi)/8 i) = cos( (3pi)/8) + i sin ((3pi)/8) #

#e^(pi/4 i) -e^((3pi)/8 i) = [cos (pi/4) +isin(pi/4)] - [cos( (3pi)/8) + i sin ((3pi)/8)]#

#e^(pi/4 i) -e^((3pi)/8 i) = [cos (pi/4) -cos( (3pi)/8)] - i[ sin(pi/4)+sin ((3pi)/8)]#

#~~0.324-i0.217#