How do you evaluate # e^( ( 3 pi)/8 i) - e^( ( pi)/6 i)# using trigonometric functions?

1 Answer
Somebody N.
Apr 15, 2018

#color(blue)(~~-0.4833419716+0.4238795325i)#

Explanation:

From Euler's formula:

#e^(itheta)=cos(theta)+isin(theta)#

#e^((3pi)/8i)-e^(pi/6i)=cos((3pi)/8)+isin((3pi)/8)-[cos(pi/6)+isin(pi/6)]#

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =cos((3pi)/8)-cos(pi/6)+(sin((3pi)/8)-sin(pi/6))i#

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

# \ \ \ \ \ \ \ \ \ \ \ ~~-0.4833419716+0.4238795325i#

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