Question

How can you use trigonometric functions to simplify `13 e^( ( pi)/8 i )` into a non-exponential complex number?

Answer 1

`13 [ cos(pi/8) + isin(pi/8) ]`

Explanation:

Using `color(blue)" Euler's relation"`

which states `re^(theta i) = r(costheta + isintheta )`

here r = 13 and `theta = pi/8`

`rArr 13e^(pi/8 i) = 13 [ cos(pi/8) + isin(pi/8) ]`