How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( 5 pi)/3 i)# using trigonometric functions?

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Answer 1 · 2017-08-19T07:03:45

The answer is #=(sqrt2/2-sqrt3/2)-i(sqrt2/2+sqrt3/2)#

We apply Euler's formula

#e^(i x)=cosx +i sinx#

So,

#e^(7/4pi i)-e^(5/3pi i)=#

#cos(7/4pi)+isin(7/4pi)-cos(5/3pi)-isin(5/3pi)#

We calculate separately

#cos(7/4pi)=cos(pi+3/4pi)=cos(pi)cos(3/4pi)-sin(pi)sin(3/4pi)#

#=-1*-sqrt2/2-0=sqrt2/2#

#sin(7/4pi)=sin(pi+3/4pi)=sin(pi)cos(3/4pi)+sin(3/4pi)cos(pi)#

#=0+sqrt2/2*-1=-sqrt2/2#

#cos(5/3pi)=cos(pi+2/3pi)=cos(pi)cos(2/3pi)-sin(pi)sin(2/3pi)#

#=-1*-sqrt3/2-0=sqrt3/2#

#sin(5/3pi)=sin(pi+2/3pi)=sin(pi)cos(2/3pi)+sin(2/3pi)cos(pi)#

#=0+sqrt3/2*-1=-sqrt3/2#

#e^(7/4pi i)-e^(5/3pi i)=(sqrt2/2-sqrt3/2)-i(sqrt2/2+sqrt3/2)#