Paragraph: Let f(x)=[x] , where [.] is the greatest integer...

$$\begin{aligned} & (f {}^{\circ} f)(x)=[[x]] \text{ and } f(x)=[x] \\ & \text{ Suppose } x=0.2 \\ & \Rightarrow(f {}^{\circ} f)(x)=[[0.2]]=[0]=0 \\ & f(x)=[0.2]=0 \\ & \Rightarrow f {}^{\circ} f(x)=f(x) \end{aligned}$$ Now, $(\operatorname{gog})(x)=\sin \sin x$ and $g(x)=\sin x$ At $x=0$$\begin{aligned} & (\operatorname{gog})(x)=\sin \sin 0=\sin 0=0 \\ & g(x)=\sin 0=0 \\ & \Rightarrow(\operatorname{gog})(x)=g(x) \text{ at } x=0 \end{aligned}$and this is true for $\mathrm{x}=\mathrm{np} , where \mathrm{n}=$0,1,2,3,4 \ldots \ldots$.\begin{aligned} & \because(f {}^{\circ} g)(x)=[\sin x] \\ & (g o(f {}^{\circ} g))(x)=\sin[\sin x] \\ & \because \sin x \text{ has value from }-1 \text{ to } 1 \\ & \text{ If }-1 \leq \sin x < 0 \\ & (g o(f {}^{\circ} g))(x)=\sin(-1)-\sin(-1) \\ & \text{ If } 0 \leq \sin x < 1 \\ & \therefore(\operatorname{go}(f {}^{\circ} g))(x)=\sin(0)=\sin 0 \text{ and } \\ & \text{ If } \sin x=1 \\ & \therefore(\operatorname{go}(f {}^{\circ} g))(x)=\sin(1)=\sin 1 \end{aligned}$$$