If `sinalpha, sin^2alpha, 1 , sin^4alpha and sin^6alpha` are in A.P., where `-pi < alpha < pi,` then `alpha` lies in the interval

Options

A. `(-pi//2,pi//2)`
B. `(-pi//3,pi//3)`
C. `(-pi//6,pi//6)`
D. none of these

Explanation

To solve the problem, we need to determine the values of \( \alpha \) for which the terms \( \sin \alpha, \sin^2 \alpha, 1, \sin^4 \alpha, \sin^6 \alpha \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
For five terms \( a, b, c, d, e \) to be in A.P., the following conditions must hold:
- \( 2b = a + c \)
- \( 2d = b + c \)
- \( 2e = d + c \)

We will first check the condition for the first three terms \( \sin \alpha, \sin^2 \alpha, 1 \).

2. **Setting Up the Equation**:
From the A.P. condition for the first three terms:
\[ 2 \sin^2 \alpha = \sin \alpha + 1 \]
Rearranging gives:
\[ 2 \sin^2 \alpha - \sin \alpha - 1 = 0 \]

3. **Solving the Quadratic Equation**:
We can solve the quadratic equation \( 2 \sin^2 \alpha - \sin \alpha - 1 = 0 \) using the quadratic formula:
\[ \sin \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -1, c = -1 \):
\[ \sin \alpha = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \]
This gives two solutions:
\[ \sin \alpha = 1 \quad \text{or} \quad \sin \alpha = -\frac{1}{2} \]

4. **Finding Values of \( \alpha \)**:
- For \( \sin \alpha = 1 \):
\[ \alpha = \frac{\pi}{2} \]
- For \( \sin \alpha = -\frac{1}{2} \):
\[ \alpha = -\frac{\pi}{6} \quad \text{or} \quad \alpha = \frac{7\pi}{6} \text{ (not in the range of } -\pi < \alpha < \pi\text{)} \]

5. **Verifying A.P. Condition**:
- For \( \alpha = \frac{\pi}{2} \):
\[ \sin \alpha = 1, \quad \sin^2 \alpha = 1, \quad 1, \quad \sin^4 \alpha = 1, \quad \sin^6 \alpha = 1 \]
All terms are equal, hence they are in A.P.
- For \( \alpha = -\frac{\pi}{6} \):
\[ \sin \left(-\frac{\pi}{6}\right) = -\frac{1}{2}, \quad \sin^2 \left(-\frac{\pi}{6}\right) = \frac{1}{4}, \quad 1, \quad \sin^4 \left(-\frac{\pi}{6}\right) = \frac{1}{16}, \quad \sin^6 \left(-\frac{\pi}{6}\right) = \frac{1}{64} \]
The terms are \( -\frac{1}{2}, \frac{1}{4}, 1, \frac{1}{16}, \frac{1}{64} \). These terms do not form an A.P.

6. **Conclusion**:
The only valid value for \( \alpha \) that satisfies the A.P. condition is \( \alpha = \frac{\pi}{2} \). However, since the problem asks for intervals and \( \frac{\pi}{2} \) is not included in any interval, we conclude that:

**The answer is that \( \alpha \) lies in the interval: None of these.**