Question

How do you evaluate `e^( ( 23 pi)/12 i) - e^( ( 17 pi)/12 i)` using trigonometric functions?

Answer 1

Possibly it is `(cos alpha+sin alpha)+i(cos alpha -sin alpha)` where `alpha= pi/12`

Explanation:

Noting that `e^((24pi i)/12)=1` and `e^((18pi i)/12)=-i` we take out the common factor `e^((-pi i)/12)`:

`e^((23pi)/12i)-e^((27pi)/12 i)`
`=e^((24-1)/12 pi i)-e^((18-1)/12 pi i)`
`=e^(-pi/12 i)(e^(24/12pi i) - e^(18/12pi i))`
`=e^(-pi/12 i)(1-i))`
`=(cos (pi/12) + sin (pi/12)) + i (cos (pi/12) - sin (pi /12))`
or something like that.