Consider the function `f(theta)`=`4(sin^(2) theta + cos^(4) theta)`
what is the maxium value of the funciton `f(theta)`?

Options

A. 1
B. 2
C. 2
D. 4

Explanation

To find the maximum value of the function \( f(\theta) = 4 \sin^2 \theta + \cos^4 \theta \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(\theta) = 4 \sin^2 \theta + \cos^4 \theta \]
We know that \( \cos^4 \theta = (\cos^2 \theta)^2 \). Let’s express \( \cos^2 \theta \) in terms of \( \sin^2 \theta \):
\[ \cos^2 \theta = 1 - \sin^2 \theta \]
Thus,
\[ \cos^4 \theta = (1 - \sin^2 \theta)^2 \]

### Step 2: Substitute and simplify
Substituting this back into the function gives:
\[ f(\theta) = 4 \sin^2 \theta + (1 - \sin^2 \theta)^2 \]
Expanding \( (1 - \sin^2 \theta)^2 \):
\[ (1 - \sin^2 \theta)^2 = 1 - 2\sin^2 \theta + \sin^4 \theta \]
So now we can rewrite \( f(\theta) \):
\[ f(\theta) = 4 \sin^2 \theta + 1 - 2 \sin^2 \theta + \sin^4 \theta \]
Combining like terms:
\[ f(\theta) = \sin^4 \theta + 2 \sin^2 \theta + 1 \]

### Step 3: Let \( x = \sin^2 \theta \)
Let \( x = \sin^2 \theta \). Then \( f(x) = x^2 + 2x + 1 \). This is a quadratic function in terms of \( x \):
\[ f(x) = (x + 1)^2 \]

### Step 4: Find the maximum value
Since \( x = \sin^2 \theta \) can take values from \( 0 \) to \( 1 \), we evaluate \( f(x) \) at the endpoints:
- When \( x = 0 \):
\[ f(0) = (0 + 1)^2 = 1 \]
- When \( x = 1 \):
\[ f(1) = (1 + 1)^2 = 4 \]

### Conclusion
The maximum value of \( f(\theta) \) occurs when \( \sin^2 \theta = 1 \) (i.e., \( \theta = \frac{\pi}{2} + n\pi \) for integers \( n \)), and the maximum value is:
\[ \boxed{4} \]