According to the Weierstrass factorization theorem, the sine function can be represented as a product of its factors:
$$\sin(x)-0=(x) \left(1 - \frac{x}{\pi} \right) \left(1 + \frac{x}{\pi} \right) \left(1 - \frac{x}{2\pi} \right) \left(1 + \frac{x}{2\pi} \right) \left(1 - \frac{x}{3\pi} \right) \left(1 + \frac{x}{3\pi} \right) \cdots$$
Would it be wrong to extend this for any $\sin(x)-k=0$??
Is it valid to represent $$\sin(x)-\frac{1}{\sqrt{2}}=(\frac{-1}{\sqrt{2}}) \left(1 - \frac{4x}{\pi} \right) \left(1 - \frac{4x}{3\pi} \right) \left(1 + \frac{4x}{5\pi} \right) \left(1 + \frac{4x}{7\pi} \right) \cdots??$$
(Since, the factors of $\sin(x)-\dfrac{1}{\sqrt{2}}$ are: $\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{-5\pi}{4},\dfrac{-7\pi}{4},\cdots$)
Or that, $$\sin(x)-\cos(x)=(-1) \left(1 - \frac{4x}{\pi} \right) \left(1 + \frac{4x}{3\pi} \right) \left(1 - \frac{4x}{5\pi} \right) \left(1 + \frac{4x}{7\pi} \right) \cdots??$$
(Since, the factors of $\sin(x)-\cos(x)$ are: $\dfrac{\pi}{4},\dfrac{-3\pi}{4},\dfrac{5\pi}{4},\dfrac{-7\pi}{4},\cdots$)
The Weierstraß product theorem is more complicated, it says, that for a set of set zeros and multiplicities ${a_k,n_k}$, without a finite limit point of the postions $a_k$, there exists analytic functions
$$F_0(z) = (z-a_0)^{n_0}\prod_{m=1}^\infty \left( (1-\frac{z-a_0}{a_m-a_0}) \ e^{\text{A}_m}\right)^{n_m} $$
with
$$\text{A}_m = \sum_{n=0}^m \frac{1}{k_n} \left( \frac{z-a_0}{a_m-a_0} \right)^{k_n} $$
For an appropriate choice of the set 0f $k$. Any two such functions differ by an entier function.
With all multiplicities equal to 1 the exponents $n_m$ disapear.