If ` cos alpha + cos beta + cos gamma = 0 = sin alpha + sin beta + sin gamma ` then
` cos ( beta + gamma) + cos ( gamma + alpha) + cos ( alpha + beta) = 0 `
sin ` ( beta + gamma ) + sin ( gamma + alpha ) + sin ( alpha + beta ) = 0 `

Answer

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

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

We need to prove that:

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

### Step 1: Represent the angles using complex exponentials

We can represent the angles using Euler's formula:

\[ e^{i\alpha} = \cos \alpha + i \sin \alpha \]
\[ e^{i\beta} = \cos \beta + i \sin \beta \]
\[ e^{i\gamma} = \cos \gamma + i \sin \gamma \]

### Step 2: Sum the complex exponentials

Now, we can write:

\[ e^{i\alpha} + e^{i\beta} + e^{i\gamma} = 0 \]

This means that the sum of the complex numbers corresponding to the angles is zero.

### Step 3: Use the property of complex numbers

If \( e^{i\alpha} + e^{i\beta} + e^{i\gamma} = 0 \), then the vectors represented by these complex numbers form a closed triangle in the complex plane. This implies that the angles \( \alpha, \beta, \gamma \) are such that they can be represented as vectors that sum to zero.

### Step 4: Find the sum of cosines

Using the identity for the cosine of sums, we can express:

\[ \cos(\beta + \gamma) = \cos \beta \cos \gamma - \sin \beta \sin \gamma \]
\[ \cos(\gamma + \alpha) = \cos \gamma \cos \alpha - \sin \gamma \sin \alpha \]
\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

Now, summing these:

\[ \cos(\beta + \gamma) + \cos(\gamma + \alpha) + \cos(\alpha + \beta) \]

Substituting the identities:

\[ = (\cos \beta \cos \gamma - \sin \beta \sin \gamma) + (\cos \gamma \cos \alpha - \sin \gamma \sin \alpha) + (\cos \alpha \cos \beta - \sin \alpha \sin \beta) \]

### Step 5: Group terms

Grouping the cosine terms and sine terms, we have:

\[ = (\cos \beta \cos \gamma + \cos \gamma \cos \alpha + \cos \alpha \cos \beta) - (\sin \beta \sin \gamma + \sin \gamma \sin \alpha + \sin \alpha \sin \beta) \]

### Step 6: Use the given conditions

From the conditions \( \cos \alpha + \cos \beta + \cos \gamma = 0 \) and \( \sin \alpha + \sin \beta + \sin \gamma = 0 \), we can conclude that the sums of the products of the cosines and sines will also yield zero.

Thus, we have:

\[ \cos(\beta + \gamma) + \cos(\gamma + \alpha) + \cos(\alpha + \beta) = 0 \]

### Step 7: Find the sum of sines

Similarly, for the sine terms:

\[ \sin(\beta + \gamma) = \sin \beta \cos \gamma + \cos \beta \sin \gamma \]
\[ \sin(\gamma + \alpha) = \sin \gamma \cos \alpha + \cos \gamma \sin \alpha \]
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]

Summing these gives:

\[ \sin(\beta + \gamma) + \sin(\gamma + \alpha) + \sin(\alpha + \beta) = 0 \]

### Conclusion

Thus, we have shown that both conditions hold true:

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