Question

How can you use trigonometric functions to simplify `9 e^( ( 11 pi)/6 i )` into a non-exponential complex number?

Answer 1

`9cos(pi/6)-i9sin(pi/6)`

Explanation:

Euler's formula states:

`e^{ix}=\cos x+ i \sin x`

Hence,

`e^{i(11pi)/6}=cos( (11pi)/6)+ i sin((11pi)/6) = cos(pi/6)-isin(pi/6)`

`:.9e^{i(11pi)/6}=9cos(pi/6)-i9sin(pi/6)`