If `A = sin^4 theta + cos^4 theta,` then

Options

A. `0 lt A lt 1/2`
B. `1/2 le A le 1`
C. `1 lt A le 3/2`
D. `3/2 le A le 2`

Answer

To solve the problem where \( A = \sin^4 \theta + \cos^4 \theta \), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite \( A \) using the identity for the sum of squares:
\[ A = \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta)^2 + (\cos^2 \theta)^2 \]

### Step 2: Use the identity for squares
We can use the identity \( a^2 + b^2 = (a + b)^2 - 2ab \):
\[ A = (\sin^2 \theta + \cos^2 \theta)^2 - 2(\sin^2 \theta \cos^2 \theta) \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ A = 1^2 - 2(\sin^2 \theta \cos^2 \theta) = 1 - 2\sin^2 \theta \cos^2 \theta \]

### Step 3: Use the double angle identity
We know that \( \sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2(2\theta) \):
\[ A = 1 - 2 \cdot \frac{1}{4} \sin^2(2\theta) = 1 - \frac{1}{2} \sin^2(2\theta) \]

### Step 4: Determine the range of \( A \)
The function \( \sin^2(2\theta) \) varies between 0 and 1. Therefore:
- The maximum value of \( A \) occurs when \( \sin^2(2\theta) = 0 \):
\[ A_{\text{max}} = 1 - \frac{1}{2} \cdot 0 = 1 \]
- The minimum value of \( A \) occurs when \( \sin^2(2\theta) = 1 \):
\[ A_{\text{min}} = 1 - \frac{1}{2} \cdot 1 = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 5: Conclusion
Thus, the interval for \( A \) is:
\[ \frac{1}{2} \leq A \leq 1 \]

### Final Answer
The value of \( A \) varies in the interval \( \left[\frac{1}{2}, 1\right] \).

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