Question

PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS

Answer 1

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

$(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}$

Then

$\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)$

So we end up with

$2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)$

Answer 2

Answer:

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

Step-by-step 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)

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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.