Part of it is that Newtonian mechanics is described in terms of calculus.
When we consider vibrational motions, we're talking about some particle that tends to not be displaced from some equilibrium position. That is, the force on the particle, at displacement $x$, $F(x)$, is equal to some function of displacement $x$, $g(x)$.
There are two ways calculus gets involved here. Firstly, $F=ma$, and $a$, acceleration, is a "rate of change" and therefore a calculus concept. So we have $ma(x)=g(x)$.
Now, dealing with a general function $g$ is too difficult - we won't get anywhere with it. So how can we proceed in the most general way? One fruitful method is to do a Taylor expansion. $g(x)=g(0)+g'(0) x+\frac{1}{2} g''(0) x^2+\frac{1}{3!} g^{(3)}(0)x^3+\cdots$, where these are the $g^{(n)}(x)$ is the nth derivative of g at point x.
If we want $x=0$ to be an equilibrium position, we must have $g(0)=0$ - there isn't any force on the particle at equilibrium. If we want it to be a stable equilibrium that will tend to turn back to its original position, we must have $g'(0)<0$. All other derivatives are fair game. Writing $-k=g'(0)$:
$$m a(x)=-k x+\frac{1}{2} g''(0) x^2+\frac{1}{3!} g^{(3)}(0)x^3+\cdots$$
as is so useful in physics, we now suppose that $x$ is small, so that $x^2$ is very small, and $x^3$ is even smaller. That is, we ignore all powers of $x$ greater than one. We wind up with:
$$m a(x)=-k x$$
Hooke's law. The solution to this equation is sinusoidal, always. (that is, it can be written in the form $x=a \cos(\omega t-\varphi)$)
So it is inevitable that, with these definitions of "stable equilibrium", the resulting vibrational pattern at small amplitudes will be sinusoidal. Always. That's what makes $\cos$ and $\sin$ special from a physical point of view.
(of course, we've also tacitly assumed that $g$ is a nice function that is nice and smooth and differentiable, but one generally does that when working on Newtonian style problems)