Let $f(x)=\sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\right)$ for all $x \in R$ and $g(x)=\frac{\pi}{2}\sin x$ for all $x \in R$. Let $(fog)(x)$ denote $f(g(x))$ and $(gof)(x)$ denote $g(f(x))$. Then which of the following is (are) true?

$\text{Given: } f(x) = \sin\left(\dfrac{\pi}{6}\sin\left(\dfrac{\pi}{2}\sin x\right)\right)$ $g(x) = \dfrac{\pi}{2}\sin x,\qquad x \in \mathbb{R}$ $\text{To find: which of } (1),\ (2),\ (3),\ (4) \text{ are true}$ $\text{Plan: range}(g) \to \text{range}(f) \to \text{range}(f\circ g) \to \text{limit} \to (g\circ f)$ $g(x) = \dfrac{\pi}{2}\sin x \Rightarrow \text{range}(g) = \left[-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right]$ $\dfrac{\pi}{2}\sin x \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right] \Rightarrow \sin\left(\dfrac{\pi}{2}\sin x\right) \in [-1,1]$ $\dfrac{\pi}{6}\sin\left(\dfrac{\pi}{2}\sin x\right) \in \left[-\dfrac{\pi}{6},\dfrac{\pi}{6}\right]$ $f(x) \in \left[-\sin\dfrac{\pi}{6},\,\sin\dfrac{\pi}{6}\right] = \left[-\dfrac{1}{2},\dfrac{1}{2}\right] \Rightarrow (1)\ \text{TRUE}$ $f\circ g = f(g(x)),\qquad g \text{ is onto } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ $\Rightarrow \text{Range}(f\circ g) = \left[-\dfrac{1}{2},\dfrac{1}{2}\right]$ $\Rightarrow (2)\ \text{TRUE}$ $x\to 0:\ \sin x \approx x \Rightarrow \dfrac{\pi}{2}\sin x \approx \dfrac{\pi}{2}x$ $f(x) \approx \sin\left(\dfrac{\pi}{6}\cdot\dfrac{\pi}{2}x\right) \approx \dfrac{\pi^2 x}{12}$ $\dfrac{f(x)}{g(x)} \approx \dfrac{\pi^2 x/12}{(\pi/2)\,x} = \dfrac{\pi}{6} \Rightarrow (3)\ \text{TRUE}$ $(g\circ f)(x) = \dfrac{\pi}{2}\sin(f(x)),\qquad f(x) \in \left[-\dfrac{1}{2},\dfrac{1}{2}\right]$ $\max (g\circ f) = \dfrac{\pi}{2}\sin\dfrac{1}{2} \approx 0.75 < 1$ $\therefore\ (g\circ f)(x) = 1 \text{ is impossible} \Rightarrow (4)\ \text{FALSE}$ $(1),\ (2),\ (3)\ \text{are TRUE};\quad (4)\ \text{is FALSE}$ $\boxed{\text{Answer: } (1),\ (2),\ (3)}$