Transforming position function into a sinusoidal function?

Question

Im trying to transform the position function $x(t)=3\sin(\pi t)+4\cos(\pi t)$, into a sinusoidal function of the form $x(t)=A\sin(\omega t+\phi )$.

Im trying to follow the steps over at this question: What is the phase shift of a sinusoidal function?

Why is it necessary to divide the position function by $\sqrt{3^2+4^2}=5?$

That and the fact that the question im trying to follow uses the sinusoidal form $f(x) = A \sin(k(\delta + x))$ instead , makes me unsure of what to do?

Answer 1

You can expand the function $A\sin(\omega t+ \phi)$ using the trigonometric sum of angle identity,

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

Equating the right hand side to $3\sin(\pi t)+4\cos(\pi t)$, we have $A\cos(\phi)=3$, $A\sin(\phi)=4$ and $\omega = \pi$

So $A^2\cos^2(\phi)+A^2\sin^2(\phi)=3^2+4^2\Rightarrow A=\sqrt{3^2+4^2}=5$ and $\phi=\arctan\left(\frac{4}{3}\right)$

Putting everything together, we have

$$3\sin(\pi t)+4\cos(\pi t)=5\sin\left(\pi t+\arctan\left(\frac{4}{3}\right)\right)$$

Thus, the phase shift is $\arctan\left(\frac{4}{3}\right)$ - this term is independent of the $t$ (time) parameter, and hence is fixed.

Answer 2

You can look at it like this:

$$3\sin(\pi t) + 4\cos(\pi t)=\Im\{3e^{i\pi t}+4i e^{i\pi t}\}=\Im\{(3+4i)e^{i\pi t}\}\\ \Im\{\sqrt{3^2+4^2}e^{i\arctan(4/3)}e^{i\pi t}\}=5\sin[\pi t + \arctan(4/3)]$$