214k views
3 votes
PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS

PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS-example-1
User Prabhuraj
by
5.1k points

2 Answers

5 votes

You want to end up with
A\sin(\omega t+\phi). Expand this using the angle sum identity for sine:


A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi

We want this to line up with
2\sin(4\pi t)+5\cos(4\pi t). Right away, we know
\omega=4\pi.

We also need to have


\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}

Recall that
\sin^2x+\cos^2x=1 for all
x; this means


(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=√(29)

Then


\begin{cases}\cos\phi=\frac2{√(29)}\\\sin\phi=\frac5{√(29)}\end{cases}\implies\tan\phi=(\sin\phi)/(\cos\phi)=\frac52\implies\phi=\tan^(-1)\left(\frac52\right)

So we end up with


2\sin(4\pi t)+5\cos(4\pi t)=√(29)\sin\left(4\pi t+\tan^(-1)\left(\frac52\right)\right)

User Muhammed Bhikha
by
5.4k points
4 votes

Answer:

  • y(t) = √29·sin(4πt +1.1903)
  • amplitude: √29
  • angular frequency: 4π
  • phase shift: 1.1903 radians

Explanation:

In the form ...

y(t) = Asin(ωt +φ)

you have ...

  • Amplitude = A
  • angular frequency = ω
  • phase shift = φ

The translation from ...

y(t) = 2sin(4πt) +5cos(4πt)

is ...

A = √(2² +5²) = √29 . . . . the amplitude

ω = 4π . . . . the angular frequency in radians per second

φ = arctan(5/2) ≈ 1.1903 . . . . radians phase shift

Then, ...

y(t) = √29·sin(4πt +1.1903)

_____

Comment on the conversion

You will notice we used "2" and "5" to find the amplitude and phase shift. In the generic case, these are "coefficient of sin( )" and "coefficient of cos( )". When determining phase shift, pay attention to whether your calculator is giving you degrees or radians. (Set the mode to what you want.)

If you have a negative coefficient for sin( ), you will need to add 180° (π radians) to the phase shift value given by the arctan( ) function.

User Mavnn
by
4.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.