Let $f, g, h$ be real functions given by $f(x)=sin x, g(x)=2 x$ and $h(x)=cos x$. Prove that fog $=g circ(f h)$.

We know that $f: R \rightarrow[-1,1]$ and $g: R \rightarrow R$

Clearly, the range of $g$ is a subset of the domain of $f$.

$f o g: R \rightarrow R$

Now, $(f h)(x)=f(x) h(x)=(\sin x)(\cos x)=\frac{1}{2} \sin (2 x)$

Domain of $f h$ is $R$.

Since range of $\sin x$ is $[-1,1]$,

$-1 \leq \sin 2 x \leq 1$

$\Rightarrow \frac{-1}{2} \leq \frac{\sin x}{2} \leq \frac{1}{2}$

Range of $f h=\left[\frac{-1}{2}, \frac{1}{2}\right]$

So, $(f h): R \rightarrow\left[\frac{-1}{2}, \frac{1}{2}\right]$

Clearly, range of $f h$ is a subset of $g$.

$\Rightarrow g o(f h): R \rightarrow R$

$\Rightarrow$ domains of $f o g$ and $g o(f h)$ are the same.

So, $(f o g)(x)=f(g(x))=f(2 x)=\sin (2 x)$

and $(g o(f h))(x)=g((f h)(x))=g(\sin x \cos x)=2 \sin x \cos x=\sin (2 x)$

$\Rightarrow(f o g)(x)=(g o(f h))(x), \forall x \in R$

Hence, $f o g=g o(f h)$