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

1 Answer
Feb 23, 2018

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

Explanation:

"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)