Question

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

Answer 1

$$\begin{gathered} \text{We know that}f:R\to \left[-1,1\right]\text{and}g:R\to R \\ \text{Clearly, the range of}\mathit{\text{g}}\text{is a subset of the domain of}\mathit{\text{f}}\text{.} \\ fog:R\to R \\ \text{Now,}\left(fh\right)\left(x\right)\text{=}f\left(x\right)h\left(x\right)=\left(\sin x\right)\left(\cos x\right)=\frac{1}{2}\sin \left(2x\right) \\ \text{Domain of}fh\text{is}R\text{.} \\ \text{Since range of sin}\mathit{\text{x}}\text{is [-1,1],} \\ -1\le \sin 2x\le 1 \\ \Rightarrow \frac{-1}{2}\le \frac{\sin x}{2}\le \frac{1}{2} \\ \text{Range of}fh\text{=}\left[\frac{-1}{2},\frac{1}{2}\right] \\ \text{So,}\left(fh\right)\text{:}R\to \left[\frac{-1}{2},\frac{1}{2}\right] \\ \mathrm{Clearly,\; range\; of}fhis\; a\; subset\; ofg. \\ \text{⇒}go\left(fh\right):R\to R \\ \text{⇒domains of}fog\text{and}go\left(fh\right)\text{are the same.} \\ \mathrm{So},\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)=f\left(2x\right)=\sin \left(2x\right) \\ \mathrm{and}\left(go\left(fh\right)\right)\left(x\right)=g\left(\left(fh\right)\left(x\right)\right)=g\left(\sin x\cos x\right)=2\sin x\cos x=\sin \left(2x\right) \\ \text{⇒}\left(fog\right)\left(x\right)=\left(go\left(fh\right)\right)\left(x\right),\forall x\in R \\ \mathrm{Hence},fog=go\left(fh\right) \end{gathered}$$