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

1 Answer
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