If `cosalpha+cosbeta=0=sinalpha+sinbeta`, then `cos2alpha+cos2beta=?`

Options

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

Answer

To solve the problem where \( \cos \alpha + \cos \beta = 0 \) and \( \sin \alpha + \sin \beta = 0 \), we need to find \( \cos 2\alpha + \cos 2\beta \).

### Step-by-Step Solution:

1. **Given Equations**:
We start with the equations:
\[ \cos \alpha + \cos \beta = 0 \]
\[ \sin \alpha + \sin \beta = 0 \]

2. **Rearranging the Equations**:
From \( \cos \alpha + \cos \beta = 0 \), we can express \( \cos \beta \) as:
\[ \cos \beta = -\cos \alpha \]
From \( \sin \alpha + \sin \beta = 0 \), we can express \( \sin \beta \) as:
\[ \sin \beta = -\sin \alpha \]

3. **Using the Pythagorean Identity**:
We know that:
\[ \cos^2 \alpha + \sin^2 \alpha = 1 \]
Therefore, substituting \( \cos \beta \) and \( \sin \beta \):
\[ \cos^2 \beta + \sin^2 \beta = \cos^2 \alpha + \sin^2 \alpha = 1 \]
This confirms that \( \alpha \) and \( \beta \) are supplementary angles.

4. **Finding \( \cos 2\alpha + \cos 2\beta \)**:
We can use the double angle formula:
\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta \]
Thus, we can write:
\[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \]
\[ \cos 2\beta = \cos^2 \beta - \sin^2 \beta \]

5. **Substituting for \( \cos^2 \beta \) and \( \sin^2 \beta \)**:
Since \( \cos \beta = -\cos \alpha \) and \( \sin \beta = -\sin \alpha \):
\[ \cos^2 \beta = \cos^2 \alpha \]
\[ \sin^2 \beta = \sin^2 \alpha \]

6. **Combining the Results**:
Now we can combine:
\[ \cos 2\alpha + \cos 2\beta = (\cos^2 \alpha - \sin^2 \alpha) + (\cos^2 \beta - \sin^2 \beta) \]
\[ = (\cos^2 \alpha - \sin^2 \alpha) + (\cos^2 \alpha - \sin^2 \alpha) \]
\[ = 2(\cos^2 \alpha - \sin^2 \alpha) \]

7. **Using the Cosine of Sum Formula**:
We can express this as:
\[ = 2 \cos(\alpha + \beta) \]
Since \( \alpha + \beta = 180^\circ \) (or \( \pi \) radians), we have:
\[ \cos(\alpha + \beta) = \cos 180^\circ = -1 \]
Therefore:
\[ \cos 2\alpha + \cos 2\beta = 2 \cdot (-1) = -2 \]

### Final Answer:
\[ \cos 2\alpha + \cos 2\beta = -2 \]