Question

How do you evaluate `e^( (3 pi)/2 i) - e^( ( 5 pi)/6 i)` using trigonometric functions?

Answer 1

`e^((3pi)/2 i) - e^((5pi)/6 i) = (sqrt3-3i)/2`

Explanation:

Recall the identity :

`e^(color(red)idelta) = cos delta + color(red)isindelta`

Given the values of `delta`, we have :

`e^((3pi)/2i) = cos((3pi)/2)+isin((3pi)/2)`

`e^((5pi)/6i)= cos((5pi)/6)+isin((5pi)/6)`

These are fairly small values of `delta`, so you can figure them out using the unit circle and properties of the trigonometric functions.

`cos((3pi)/2) = 0`
`sin((3pi)/2) = -1`

For `delta = (5pi)/6`, we have to use the equalities `sin(color(red)pi-delta) = sin delta` and `cos(color(red)pi-delta) = color(red)-cos delta`:

`sin((5pi)/6) = sin(color(red)pi-(5pi)/6) = sin (pi/6) = 1/2`

`cos((5pi)/6) = color(red)-cos(pi/6) = - sqrt3/2`

Plugging in these values for `e^((3pi)/2i) - e^((5pi)/6i)`, we get :

`e^((3pi)/2i) - e^((5pi)/6i) = 0 + i(-1) - (-sqrt3/2)-i(1/2)`

`color(red)(e^((3pi)/2i) - e^((5pi)/6i) = (sqrt3-3i)/2)`.