If ` cos alpha + cos beta = 0 = sin alpha + sin beta,` then value of `cos 2 alpha + cos 2 beta` is

Options

A. `-2 sin (alpha + beta)`
B. `2 cos ( alpha + beta)`
C. `2 sin (alpha - beta)`
D. `- 2 cos (alpha + beta)`

Answer

To solve the problem, we start with the given equations:

1. \( \cos \alpha + \cos \beta = 0 \)
2. \( \sin \alpha + \sin \beta = 0 \)

From these equations, we will find the value of \( \cos 2\alpha + \cos 2\beta \).

### Step 1: Analyze the first equation
From \( \cos \alpha + \cos \beta = 0 \), we can express \( \cos \beta \) in terms of \( \cos \alpha \):
\[ \cos \beta = -\cos \alpha \]

### Step 2: Analyze the second equation
From \( \sin \alpha + \sin \beta = 0 \), we can express \( \sin \beta \) in terms of \( \sin \alpha \):
\[ \sin \beta = -\sin \alpha \]

### Step 3: Use the Pythagorean identity
Using the identities for cosine and sine, we can derive:
\[ \cos^2 \alpha + \sin^2 \alpha = 1 \]
Substituting for \( \cos \beta \) and \( \sin \beta \):
\[ \cos^2 \beta + \sin^2 \beta = \cos^2 \alpha + \sin^2 \alpha = 1 \]

### Step 4: Find \( \cos 2\alpha + \cos 2\beta \)
Using the double angle formula:
\[ \cos 2\alpha = 2\cos^2 \alpha - 1 \]
\[ \cos 2\beta = 2\cos^2 \beta - 1 \]
Substituting \( \cos \beta = -\cos \alpha \):
\[ \cos 2\beta = 2(-\cos \alpha)^2 - 1 = 2\cos^2 \alpha - 1 \]
Thus, we have:
\[ \cos 2\alpha + \cos 2\beta = (2\cos^2 \alpha - 1) + (2\cos^2 \alpha - 1) \]
\[ = 4\cos^2 \alpha - 2 \]

### Step 5: Substitute \( \cos^2 \alpha \)
Using \( \cos^2 \alpha + \sin^2 \alpha = 1 \), we can express \( \cos^2 \alpha \) in terms of \( \sin^2 \alpha \):
\[ \cos^2 \alpha = 1 - \sin^2 \alpha \]

### Step 6: Substitute back into the equation
Now substituting back:
\[ \cos 2\alpha + \cos 2\beta = 4(1 - \sin^2 \alpha) - 2 \]
\[ = 4 - 4\sin^2 \alpha - 2 = 2 - 4\sin^2 \alpha \]

### Final Result
Since \( \sin^2 \alpha + \sin^2 \beta = 0 \) implies \( \sin^2 \alpha = 0 \) (because both sine terms must be equal and opposite), we find:
\[ \cos 2\alpha + \cos 2\beta = 2 - 4(0) = 2 \]

Thus, the value of \( \cos 2\alpha + \cos 2\beta \) is:
\[ \boxed{2} \]