How do you evaluate #e^( ( 13 pi)/8 i) - e^( ( 7 pi)/4 i)# using trigonometric functions?

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Answer 1 · 2018-02-23T12:56:47

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

#"Reminder : "#

Euler's relation

#e^(itheta)=costheta+isintheta#

Here, we have

#z=e^(13/8pi)-e^(7/4pi)=cos(13/8pi)+isin(13/8pi)-cos(7/4pi)-isin(7/4pi)#

Therefore,

#13/8pi=5/8pi+pi=1/8pi+3/2pi#

#7/4pi=3/4pi+pi#

#cos(1/4pi)=1-2sin^2(1/8pi)=2cos^2(1/8pi)-1#

#sin(1/8pi)=sqrt((1-cos(1/4pi))/2)=sqrt((1-sqrt2/2)/(2))=(sqrt(2-sqrt2))/2#

#cos(1/8pi)=sqrt((1+cos(1/4pi))/2)=sqrt((1+sqrt2/2)/(2))=(sqrt(2+sqrt2))/2##

So,
#z=cos(1/8pi+3/2pi)+isin(1/8pi+3/2pi)-cos(1/4pi+pi)-isin(1/4pi+pi)#

#cos(1/8pi+3/2pi)=cos(1/8pi)cos(3/2pi)-sin(1/8pi)sin(3/2pi)#

#=(sqrt(2+sqrt2))/2*0-(sqrt(2-sqrt2))/2*(-1)=(sqrt(2-sqrt2))/2#

#sin(1/8pi+3/2pi)=sin(1/8pi)cos(3/2pi)+cos(1/8pi)sin(3/2pi)#

#=(sqrt(2-sqrt2))/2*0+(sqrt(2+sqrt2))/2*(-1)=-(sqrt(2+sqrt2))/2#

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

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

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

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

Finally,

#z=(sqrt(2-sqrt2))/2-i(sqrt(2+sqrt2))/2-sqrt2/2+isqrt2/2#

#=((sqrt(2-sqrt2))/2-sqrt2/2)-i((sqrt(2+sqrt2))/2-sqrt2/2)#