Approximating x by a combination of sines

Expansions.
$$\sin x=x-\frac{x^3}6+\frac{x^5}{120}+o(x^5),\qquad\sin2x=2x-\frac{8x^3}6+\frac{32x^5}{120}+o(x^5).$$

Conditions. The coefficient of $x$ must be $1$ and the coefficient of $x^3$ must vanish:
$$\alpha+2\beta=1,\qquad-\frac\alpha6-\frac{8\beta}6=0\ \Longrightarrow\ \alpha=-8\beta.$$
So $-6\beta=1$, i.e. $\beta=-\frac16$ and $\alpha=\frac43$.

Error term. The coefficient of $x^5$ is $\frac{\alpha+32\beta}{120}=\frac{\frac43-\frac{16}3}{120}=-\frac1{30}$:
$$\frac43\sin x-\frac16\sin2x=x-\frac{x^5}{30}+o(x^5).$$

Remark. This is Richardson extrapolation: two approximations with the same structure of error are combined to cancel the leading error term. It is the idea behind many numerical methods (Romberg integration, for instance).