If `f:R -> R, g:R -> R` defined as `f(x) = sin x and g(x) = x^2`, then find the value of `(gof)(x) and (fog)(x) `and also prove that `gof != fog`.

Explanation

To solve the problem, we need to find the composite functions \( (g \circ f)(x) \) and \( (f \circ g)(x) \) where \( f(x) = \sin x \) and \( g(x) = x^2 \). We will also prove that these two composite functions are not equal.

### Step 1: Find \( (g \circ f)(x) \)

The composite function \( (g \circ f)(x) \) means we need to apply \( f \) first and then apply \( g \) to the result of \( f \).

1. Start with \( f(x) = \sin x \).
2. Now, substitute \( f(x) \) into \( g \):
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