Expanding the desired form, we have
[tex]A \sin(\omega t + \phi) = A \bigg(\sin(\omega t) \cos(\phi) + \cos(\omega t) \sin(\phi)\bigg)[/tex]
and matching it up with the given expression, we see that
[tex]\begin{cases} A \sin(\omega t) \cos(\phi) = 2 \sin(4\pi t) \\ A \cos(\omega t) \sin(\phi) = 5 \cos(4\pi t) \end{cases}[/tex]
A natural choice for one of the symbols is [tex]\omega = 4\pi[/tex]. Then
[tex]\begin{cases} A \cos(\phi) = 2 \\ A \sin(\phi) = 5 \end{cases}[/tex]
Use the Pythagorean identity to eliminate [tex]\phi[/tex].
[tex](A\cos(\phi))^2 + (A\sin(\phi))^2 = A^2 \cos^2(\phi) + A^2 \sin^2(\phi) = A^2 (\cos^2(\phi) + \sin^2(\phi)) = A^2[/tex]
so that
[tex]A^2 = 2^2 + 5^2 = 29 \implies A = \pm\sqrt{29}[/tex]
Use the definition of tangent to eliminate [tex]A[/tex].
[tex]\tan(\phi) = \dfrac{\sin(\phi)}{\cos(\phi)} = \dfrac{A\sin(\phi)}{\cos(\phi)}[/tex]
so that
[tex]\tan(\phi) = \dfrac52 \implies \phi = \tan^{-1}\left(\dfrac52\right)[/tex]
We end up with
[tex]y(t) = 2 \sin(4\pi t) + 5 \cos(4\pi t) = \boxed{\pm\sqrt{29} \sin\left(4\pi t + \tan^{-1}\left(\dfrac52\right)\right)}[/tex]
where
• amplitude:
[tex]|A| = \boxed{\sqrt{29}}[/tex]
• angular frequency:
[tex]\boxed{4\pi}[/tex]
• phase shift:
[tex]4\pi t + \tan^{-1}\left(\dfrac 52\right) = 4\pi \left(t + \boxed{\frac1{4\pi} \tan^{-1}\left(\frac52\right)}\,\right)[/tex]
