If `P=(1)/(2)sin^(2)theta+(1)/(3)cos^(2)theta`, then

Options

A. `(1)/(3)lePle(1)/(2)`
B. `Pge(1)/(2)`
C. `2lePle3`
D. `-(sqrt(13))/(6)lePle(sqrt(13))/(6)`

Explanation

To find the range of \( P = \frac{1}{2} \sin^2 \theta + \frac{1}{3} \cos^2 \theta \), we can follow these steps:

### Step 1: Rewrite \( P \) in terms of a single trigonometric function
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can express \( \sin^2 \theta \) in terms of \( \cos^2 \theta \):
\[ \sin^2 \theta = 1 - \cos^2 \theta \]
Substituting this into the expression for \( P \):
\[ P = \frac{1}{2}(1 - \cos^2 \theta) + \frac{1}{3} \cos^2 \theta \]

### Step 2: Simplify the expression
Distributing the terms:
\[ P = \frac{1}{2} - \frac{1}{2} \cos^2 \theta + \frac{1}{3} \cos^2 \theta \]
Now, combine the terms involving \( \cos^2 \theta \):
\[ P = \frac{1}{2} + \left(-\frac{1}{2} + \frac{1}{3}\right) \cos^2 \theta \]

### Step 3: Find a common denominator
To combine the coefficients of \( \cos^2 \theta \):
\[ -\frac{1}{2} + \frac{1}{3} = -\frac{3}{6} + \frac{2}{6} = -\frac{1}{6} \]
Thus, we can rewrite \( P \):
\[ P = \frac{1}{2} - \frac{1}{6} \cos^2 \theta \]

### Step 4: Determine the range of \( P \)
The term \( \cos^2 \theta \) varies between 0 and 1. Therefore:
- When \( \cos^2 \theta = 0 \):
\[ P = \frac{1}{2} - \frac{1}{6}(0) = \frac{1}{2} \]
- When \( \cos^2 \theta = 1 \):
\[ P = \frac{1}{2} - \frac{1}{6}(1) = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

### Step 5: Conclusion about the range
Thus, the range of \( P \) is from \( \frac{1}{3} \) to \( \frac{1}{2} \).

### Final Answer:
The range of \( P \) is \( \left[\frac{1}{3}, \frac{1}{2}\right] \).

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