where `f(x)=cosx,g(x)=sinx^2` ,Is fog = gof

Answer

To determine whether \( f(g(x)) = g(f(x)) \) for the functions \( f(x) = \cos x \) and \( g(x) = \sin^2 x \), we will compute \( f(g(x)) \) and \( g(f(x)) \) step by step.

### Step 1: Compute \( f(g(x)) \)

1. Start with \( g(x) \):
\[ g(x) = \sin^2 x \]

2. Substitute \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f(\sin^2 x) = \cos(\sin^2 x) \]

### Step 2: Compute \( g(f(x)) \)

1. Start with \( f(x) \):
\[ f(x) = \cos x \]

2. Substitute \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g(\cos x) = \sin^2(\cos x) \]

### Step 3: Compare \( f(g(x)) \) and \( g(f(x)) \)

Now we have:
- \( f(g(x)) = \cos(\sin^2 x) \)
- \( g(f(x)) = \sin^2(\cos x) \)

To check if they are equal, we need to see if:
\[ \cos(\sin^2 x) = \sin^2(\cos x) \]

### Conclusion

Since \( \cos(\sin^2 x) \) and \( \sin^2(\cos x) \) are not generally equal for all \( x \), we conclude that:
\[ f(g(x)) \neq g(f(x)) \]

Thus, \( f(g(x)) \) is not equal to \( g(f(x)) \).

### Final Answer

No, \( f(g(x)) \neq g(f(x)) \).

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