Question

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

Answer 1

`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`