Could someone provide additional steps to get to the answer?

[Rogatus]

Homework Statement

determine A and φ: f(x)=Acos(2t+φ)=4cos(2t+1/4π)+5sin(2t)

Homework Equations

the answer has to be in the form of Ae^φi

The Attempt at a Solution

for working towards the answer, the tekst states:

α₁=4e^1/4πi=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 4e^1/4πi 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

 

[ehild]

Homework Statement

determine A and φ: f(x)=Acos(2t+φ)=4cos(2t+1/4π)+5sin(2t)

Homework Equations

the answer has to be in the form of Ae^φi

The Attempt at a Solution

for working towards the answer, the tekst states:

α₁=4e^1/4πi=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 4e^1/4πi 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

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.

 

[Ray Vickson]

Homework Statement

determine A and φ: f(x)=Acos(2t+φ)=4cos(2t+1/4π)+5sin(2t)

Homework Equations

the answer has to be in the form of Ae^φi

The Attempt at a Solution

for working towards the answer, the tekst states:

α₁=4e^1/4πi=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 4e^1/4πi 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

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 $\sin(2t)$?