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

Hint: 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.