If `cos (theta - alpha), cos theta, cos (theta + alpha)` are in H.P. then `cos 2 theta` is equal to

Options

A. `1 - cos alpha`
B. `1 + cos alpha`
C. `1 - 2 cos alpha`
D. `1 + 2 cos alpha`

Answer

To solve the problem, we need to find the value of \( \cos 2\theta \) given that \( \cos(\theta - \alpha), \cos \theta, \cos(\theta + \alpha) \) are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding H.P.**:
If three quantities \( A, B, C \) are in H.P., then the relationship is given by:
\[ B = \frac{2AC}{A + C} \]
Here, let \( A = \cos(\theta - \alpha) \), \( B = \cos \theta \), and \( C = \cos(\theta + \alpha) \).

2. **Setting up the equation**:
According to the H.P. condition:
\[ \cos \theta = \frac{2 \cos(\theta - \alpha) \cos(\theta + \alpha)}{\cos(\theta - \alpha) + \cos(\theta + \alpha)} \]

3. **Using the cosine addition formula**:
We know that:
\[ \cos(\theta - \alpha) + \cos(\theta + \alpha) = 2 \cos \theta \cos \alpha \]
Therefore, we can substitute this into our equation:
\[ \cos \theta = \frac{2 \cos(\theta - \alpha) \cos(\theta + \alpha)}{2 \cos \theta \cos \alpha} \]

4. **Simplifying the equation**:
This simplifies to:
\[ \cos \theta \cdot 2 \cos \theta \cos \alpha = 2 \cos(\theta - \alpha) \cos(\theta + \alpha) \]
Dividing both sides by 2:
\[ \cos^2 \theta \cos \alpha = \cos(\theta - \alpha) \cos(\theta + \alpha) \]

5. **Using the product-to-sum identities**:
We can use the identity:
\[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \]
Thus:
\[ \cos(\theta - \alpha) \cos(\theta + \alpha) = \frac{1}{2} [\cos(2\theta) + \cos(0)] = \frac{1}{2} [\cos(2\theta) + 1] \]

6. **Setting the equation**:
Now we have:
\[ \cos^2 \theta \cos \alpha = \frac{1}{2} [\cos(2\theta) + 1] \]
Multiplying both sides by 2:
\[ 2 \cos^2 \theta \cos \alpha = \cos(2\theta) + 1 \]

7. **Rearranging the equation**:
Rearranging gives:
\[ \cos(2\theta) = 2 \cos^2 \theta \cos \alpha - 1 \]

8. **Using the double angle formula**:
We know that:
\[ \cos(2\theta) = 2 \cos^2 \theta - 1 \]
Therefore, we can set:
\[ 2 \cos^2 \theta - 1 = 2 \cos^2 \theta \cos \alpha - 1 \]

9. **Final simplification**:
This implies:
\[ 2 \cos^2 \theta = 2 \cos^2 \theta \cos \alpha \]
Dividing both sides by \( 2 \cos^2 \theta \) (assuming \( \cos^2 \theta \neq 0 \)):
\[ 1 = \cos \alpha \]
Thus, \( \alpha = 0 \) or \( \cos 2\theta = 1 + 2 \cos \alpha \).

### Conclusion:
Thus, the value of \( \cos 2\theta \) is:
\[ \cos 2\theta = 1 + 2 \cos \alpha \]