What is the minimum value of the function `f(theta)` ?

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the minimum 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:
\[ f(\theta) = 4 \sin^2 \theta + \cos 4\theta \]

### Step 2: Use Trigonometric Identities
We can use the double angle identity for cosine:
\[ \cos 4\theta = 1 - 2\sin^2 2\theta \]
Thus, we can rewrite \( f(\theta) \):
\[ f(\theta) = 4 \sin^2 \theta + (1 - 2\sin^2 2\theta) \]

### Step 3: Express \( \sin^2 2\theta \)
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we have:
\[ \sin^2 2\theta = (2 \sin \theta \cos \theta)^2 = 4 \sin^2 \theta \cos^2 \theta \]
Substituting this back into the function gives:
\[ f(\theta) = 4 \sin^2 \theta + 1 - 2(4 \sin^2 \theta \cos^2 \theta) \]
\[ = 4 \sin^2 \theta + 1 - 8 \sin^2 \theta \cos^2 \theta \]

### Step 4: Simplify the Function
Now we simplify:
\[ f(\theta) = 4 \sin^2 \theta + 1 - 8 \sin^2 \theta (1 - \sin^2 \theta) \]
Let \( x = \sin^2 \theta \):
\[ f(x) = 4x + 1 - 8x(1 - x) \]
\[ = 4x + 1 - 8x + 8x^2 \]
\[ = 8x^2 - 4x + 1 \]

### Step 5: Find the Minimum Value
To find the minimum value of the quadratic function \( f(x) = 8x^2 - 4x + 1 \), we can use the vertex formula \( x = -\frac{b}{2a} \):
\[ x = -\frac{-4}{2 \cdot 8} = \frac{4}{16} = \frac{1}{4} \]

### Step 6: Calculate \( f\left(\frac{1}{4}\right) \)
Now substituting \( x = \frac{1}{4} \) back into the function:
\[ f\left(\frac{1}{4}\right) = 8\left(\frac{1}{4}\right)^2 - 4\left(\frac{1}{4}\right) + 1 \]
\[ = 8 \cdot \frac{1}{16} - 1 + 1 = \frac{1}{2} - 1 + 1 = \frac{1}{2} \]

### Step 7: Conclusion
The minimum value of the function \( f(\theta) \) is:
\[ \boxed{1} \]