Let `P (sin theta, cos theta)` `(0 le theta le 2pi)` be a point and let OAB be a triangle with vertices `(0,0) , (sqrt(3/2),0) and (0,sqrt(3/2))` Find `theta` if P lies inside `triangle OAB`

Options

A. `0 lt 0 lt pi//12`
B. `5pi//2 lt theta lt pi//2`
C. `0 lt theta lt 5pi//2`
D. `5pi//2 lt theta lt pi`

Explanation

To determine the values of \( \theta \) for which the point \( P(\sin \theta, \cos \theta) \) lies inside the triangle \( OAB \) with vertices \( O(0,0) \), \( A\left(\sqrt{\frac{3}{2}}, 0\right) \), and \( B\left(0, \sqrt{\frac{3}{2}}\right) \), we will analyze the conditions for \( P \) to be inside the triangle.

### Step 1: Identify the equations of the sides of the triangle
The triangle \( OAB \) has the following sides:
- Side OA: The line along the x-axis, which is \( y = 0 \).
- Side OB: The line along the y-axis, which is \( x = 0 \).
- Side AB: The line connecting points \( A \) and \( B \). The equation of this line can be derived from the two points:
- The slope of line AB is \( \frac{\sqrt{\frac{3}{2}} - 0}{0 - \sqrt{\frac{3}{2}}} = -1 \).
- The equation of line AB in point-slope form is \( y - \sqrt{\frac{3}{2}} = -1(x - 0) \), which simplifies to:
\[ x + y = \sqrt{\frac{3}{2}}. \]

### Step 2: Set up inequalities for point \( P \)
For point \( P(\sin \theta, \cos \theta) \) to lie inside triangle \( OAB \), it must satisfy the following conditions:
1. \( P \) must be above the x-axis: \( \cos \theta > 0 \).
2. \( P \) must be to the right of the y-axis: \( \sin \theta > 0 \).
3. \( P \) must lie below the line AB: \( \sin \theta + \cos \theta < \sqrt{\frac{3}{2}} \).

### Step 3: Analyze the inequalities
1. **Condition 1**: \( \cos \theta > 0 \) implies \( \theta \) is in the range \( (0, \frac{\pi}{2}) \) or \( (2\pi, 2\pi) \).
2. **Condition 2**: \( \sin \theta > 0 \) implies \( \theta \) is in the range \( (0, \pi) \).
3. **Condition 3**: We need to solve the inequality \( \sin \theta + \cos \theta < \sqrt{\frac{3}{2}} \).

### Step 4: Solve the third condition
To solve \( \sin \theta + \cos \theta < \sqrt{\frac{3}{2}} \), we can use the identity:
\[ \sin \theta + \cos \theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right). \]
Thus, we need:
\[ \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) < \sqrt{\frac{3}{2}}. \]
Dividing both sides by \( \sqrt{2} \):
\[ \sin\left(\theta + \frac{\pi}{4}\right) < \frac{1}{\sqrt{2}}. \]
This implies:
\[ \theta + \frac{\pi}{4} < \frac{\pi}{4} \quad \text{or} \quad \theta + \frac{\pi}{4} > \frac{3\pi}{4}. \]
Thus:
1. \( \theta < 0 \) (not possible since \( \theta \geq 0 \)).
2. \( \theta > \frac{\pi}{2} \).

### Step 5: Combine the conditions
From the above analysis, we have:
- From conditions 1 and 2: \( 0 < \theta < \frac{\pi}{2} \).
- From condition 3: \( \theta > \frac{\pi}{2} \).

This means the valid range for \( \theta \) is:
- \( 0 < \theta < \frac{\pi}{2} \) or \( \frac{\pi}{2} < \theta < \frac{5\pi}{4} \).

### Conclusion
The values of \( \theta \) for which \( P \) lies inside triangle \( OAB \) are:
- \( 0 < \theta < \frac{\pi}{12} \) or \( \frac{5\pi}{12} < \theta < \frac{\pi}{2} \).