To rewrite y(t) = 2 * sin (4 pi t) + 5 * cos (4 pi t) in the form Asin(wt+phi), you must first find the amplitude. Use c1=Asin(phi) and c2=Acos(phi) along with the Pythagorean identity to solve for A (amplitude.) previous question: By plugging in for c1 and c2, the equation should shift to be: Asin(wt+phi)=Acos(phi)sin(wt)+Asin(phi)cos(wt) Use the identity sin(x+y)=sin(x)cos(y)+cos(x)sin(y)

[tex]\sin (\alpha + \beta) = \sin \alpha\cdot \cos \beta + \cos \alpha \cdot \sin \beta[/tex], [tex]\forall\,\alpha,\beta\in \mathbb{R}[/tex] (Eq. 1)

If we have formula [tex]y(t) = 2\cdot \sin (4\pi\cdot t)+5\cdot \cos (4\pi\cdot t)[/tex], by direct comparison with (Eq. 1):

[tex]A\cdot \cos \phi = 2[/tex] (Eq. 2)

[tex]A\cdot \sin \phi = 5[/tex] (Eq. 3)

Where:

[tex]A[/tex] - Amplitude, dimensionless.

[tex]\phi[/tex] - Phase angle, measured in radians.

By dividing (Eq. 3) by (Eq. 2), we get that:

[tex]\tan\phi = \frac{5}{2}[/tex]

[tex]\phi = \tan^{-1}\left(\frac{5}{2} \right)[/tex]

[tex]\phi = 0.379\pi\,rad[/tex]

From (Eq. 2) we find the amplitude:

[tex]A = \frac{2}{\cos \phi}[/tex]

[tex]A = \frac{2}{\cos 0.379\pi}[/tex]

[tex]A = 5.390[/tex]

Therefore, we find that equation in standard form is:

[tex]y(t) = A\cdot \sin (\omega\cdot t +\phi)[/tex] (Eq. 4)

Where [tex]\omega[/tex] is the angular frequency, measured in radians per second.

([tex]A = 5.390[/tex], [tex]\omega = 4\pi\,\frac{rad}{s}[/tex], [tex]\phi = 0.379\pi\,rad[/tex])

[tex]y(t) = 5.390\cdot \sin (4\pi\cdot t +0.379\pi)[/tex]