Question

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

Answer 1

According to the question, calculate the domain using the function $f\left( x \right),g\left( x \right)$ and $h\left( x \right)$ . So, that all the domains are real and hence using them we can calculate $fog$ and $go\left( {fh} \right)$ and verify that they are equal or not.

Formula used:
Here, we use the formula ${\mathop{\rm Sin}\nolimits} 2x = 2\sin x\cos x$ .

Complete step-by-step answer:
It is given that $f\left( x \right) = \sin x$, $g\left( x \right) = 2x$ and $h\left( x \right) = \cos x$.
As we know,
f: $\;R{\rm{ }} \to {\rm{ }}\left[ { - 1,\;1} \right]\;$ and g: $R{\rm{ }} \to {\rm{ }}R$
As it is clear that, the range of g is a subset of the domain of f.
So, fog: $R{\rm{ }} \to {\rm{ }}R$
Now, $\left( {fh} \right)\left( x \right){\rm{ }} = {\rm{ }}f\left( x \right){\rm{ }}h\left( x \right)\;$
Put the values of $f\left( x \right)$ and $h\left( x \right)$ in the above equation.
So, we get $= \left( {\cos x} \right)\left( {\sin x} \right)$
Multiply and divide with 2.
$= \dfrac{2}{2}\left( {\cos x} \right)\left( {\sin x} \right)$
Here, we use the identity ${\mathop{\rm Sin}\nolimits} 2x = 2\sin x\cos x$.
So, we get $= \dfrac{1}{2}\sin 2x$.
So, the domain of fh is R.
Since range of $\sin x$ is $\left[ { - 1,{\rm{ }}1} \right],{\rm{ }} - 1\; \le \;sin\;2x\; \le \;1$
On further simplifying we get,
$\Rightarrow - \dfrac{1}{2} \le sin\dfrac{x}{2}{\rm{ }} \le \dfrac{1}{2}$
Hence, the Range of fh = $\left[ { - \dfrac{1}{2},{\rm{ }}\dfrac{1}{2}} \right]$
Therefore, (fh): $R\; \to \left[ { - \dfrac{1}{2},{\rm{ }}\dfrac{1}{2}} \right]$
As it is clear from above, the range of fh is a subset of g.
So, $\Rightarrow go\left( {fh} \right):\;R\; \to \;R$
Hence, Domains of fog and go(fh) are the same.
So, here we calculate $\left( {fog} \right)\left( x \right){\rm{ }} = {\rm{ }}f\left( {g\left( x \right)} \right)\;$
By Putting the value g(x) in above we get,
$\Rightarrow f\left( {2x} \right)$
As, $f\left( x \right) = \sin x$
Therefore, $\left( {fog} \right)\left( x \right){\rm{ }} = {\rm{ }}f\left( {g\left( x \right)} \right)\;$ $\Rightarrow \sin \left( {2x} \right)$
And we also calculate $\left( {go\left( {fh} \right)} \right)\left( x \right)\; = \;g\left( {\left( {fh} \right)\left( x \right)} \right)\;$
By Putting the value $fh\left( x \right) = \sin x\cos x$in above we get,
$\Rightarrow g\left( {\sin x\cos x} \right)$
As, $g\left( x \right) = 2x$
So, we get $\Rightarrow 2\sin x\cos x$
By using the identity ${\mathop{\rm Sin}\nolimits} 2x = 2\sin x\cos x$
Therefore, \left( {go\left( {fh} \right)} \right)\left( x \right)\; = \;g\left( {\left( {fh} \right)\left( x \right)} \right)\;$$$$ \Rightarrow \sin \left( {2x} \right)
Hence, it is clear $fog\; = \;go\left( {fh} \right)$ .

Note: To solve these types of questions, we use f of g which means putting function g in function f(x). These types of problems can use chaining which means using multiple functions in functions a clear example of that is f(g((x))). We can also use the identities to solve the functions.