1. Nov 11, 2015

### Rogatus

1. The problem statement, all variables and given/known data
determine A and φ: f(x)=Acos(2t+φ)=4cos(2t+1/4π)+5sin(2t)

2. Relevant equations
the answer has to be in the form of Aeφi

3. The attempt at a solution
for working towards the answer, the tekst states:

α1=4e1/4πi=2√2+2√2i
and
α2=5e-1/2πi=-5i

I fail to see how they get 5e-1/2πi out of 5sin(2t). Also, how they reform it to -5i and how they go from 4e1/4πi to 2√2+2√2i

from then, I guess its 2√2+(2√2-5)i=3,57e-0,65i, but it's the middle part that confuses me

2. Nov 11, 2015

### ehild

You have to write all terms as cosines. sin(2t) is the cosine of what angle? Think of the definition of sine and cosine in a right triangle.

3. Nov 11, 2015

### Ray Vickson

Why don't you use the trigonometric addition laws to write both sides of the equation as constant-coefficient linear combinations of $\cos(2t)$ and $\sin(2t)$?