Question

How can you use trigonometric functions to simplify `2 e^( ( 7 pi)/12 i )` into a non-exponential complex number?

Answer 1

Explanation is given below.

Explanation:

Euler's formula `e^(i theta) = cos(theta) + i sin(theta)`

Our question `2e^((7pi)/12 i)` can be simplified to using the Euler's formula as

`2(cos((7pi)/12)+ i sin((7pi)/12))`

Now we need to evaluate this.

`cos((7pi)/12) = cos(pi/4 + pi/3)`
`cos((7pi)/12) = cos(pi/4)cos(pi/3) - sin(pi/4)sin(pi/3)`
`cos((7pi)/12) = sqrt(2)/2 * 1/2 - sqrt(2)/2 * sqrt(3)/2`
`cos((7pi)/12) = sqrt(2)/4 - sqrt(6)/4`
`cos((7pi)/12) = (sqrt(2)-sqrt(6))/4`

`sin((7pi)/12) = sin(pi/4 + pi/3)`
`sin((7pi)/12) = sin(pi/4)cos(pi/3) + cos(pi/4)sin(pi/3)`
`sin((7pi)/12) = sqrt(2)/2*1/2 +sqrt(2)/2*sqrt(3)/2`
`sin((7pi)/12) = sqrt(2)/4 + sqrt(6)/4`
`sin((7pi)/12) = (sqrt(2)+sqrt(6))/4`

The complex number would be
`2( (sqrt(2)-sqrt(6))/4 + i (sqrt(2)+sqrt(6))/4)`
`1/2 ( (sqrt(2)-sqrt(6)) + i (sqrt(2)+sqrt(6)))`

That should do for an answer, further simplification is possible depending on how the answer needs to be represented.