Could someone provide additional steps to get to the answer?

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SUMMARY

The discussion focuses on determining the amplitude (A) and phase (φ) for the function f(x) = A cos(2t + φ), specifically transforming the expression 4 cos(2t + 1/4π) + 5 sin(2t) into the required form Ae^(φi). Participants clarify that 5 sin(2t) can be expressed as 5e^(-1/2πi) = -5i, and that 4e^(1/4πi) simplifies to 2√2 + 2√2i. The confusion arises around the transformation steps and the use of trigonometric identities to combine terms.

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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 Aeφi

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
 
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Rogatus said:

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 Aeφi

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
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.
 
Rogatus said:

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 Aeφi

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

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)##?
 

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