If `cos alpha + 2 cos beta +3 cos gamma = sin alpha + 2 sin beta + 3 sin gamma y = 0`, then the value of `sin 3alpha + 8 sin 3beta + 27 sin 3gamma ` is

Options

A. `sin(a+ b+ gamma)`
B. `3 sin (alpha + beta + gamma)`
C. ` 18 sin(alpha + beta + gamma)`
D. `sin(alpha+ beta + gamma)`

Explanation

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

\[ \cos \alpha + 2 \cos \beta + 3 \cos \gamma = \sin \alpha + 2 \sin \beta + 3 \sin \gamma = 0 \]

We need to find the value of:

\[ \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma \]

### Step 1: Express the given equation in terms of complex numbers

We can express the cosine and sine functions in terms of complex exponentials:

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

Thus, we can rewrite the equation as:

\[ e^{i\alpha} + 2e^{i\beta} + 3e^{i\gamma} = 0 \]

### Step 2: Identify the real and imaginary parts

From the equation, we can separate the real and imaginary parts:

- Real part: \(\cos \alpha + 2 \cos \beta + 3 \cos \gamma = 0\)
- Imaginary part: \(\sin \alpha + 2 \sin \beta + 3 \sin \gamma = 0\)

### Step 3: Use the identity for cubes

We can use the identity for the sum of cubes:

\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) \]

In our case, let:

- \(x = \cos \alpha\)
- \(y = 2 \cos \beta\)
- \(z = 3 \cos \gamma\)

Since \(x + y + z = 0\), we can apply the identity:

\[ x^3 + y^3 + z^3 = 3xyz \]

### Step 4: Calculate \(x^3 + y^3 + z^3\)

Now, we calculate:

\[ \cos^3 \alpha + 8 \cos^3 \beta + 27 \cos^3 \gamma = 3(\cos \alpha)(2 \cos \beta)(3 \cos \gamma) \]

### Step 5: Find the value of \( \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma \)

Using the sine triple angle formula:

\[ \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta \]

We can express:

\[ \sin 3\alpha = 3 \sin \alpha - 4 \sin^3 \alpha \]
\[ \sin 3\beta = 3 \sin \beta - 4 \sin^3 \beta \]
\[ \sin 3\gamma = 3 \sin \gamma - 4 \sin^3 \gamma \]

Substituting these into our expression gives:

\[ \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma = (3 \sin \alpha + 24 \sin \beta + 81 \sin \gamma) - 4(\sin^3 \alpha + 8 \sin^3 \beta + 27 \sin^3 \gamma) \]

### Step 6: Substitute back using the earlier results

Since we know that:

\[ \sin \alpha + 2 \sin \beta + 3 \sin \gamma = 0 \]

We can conclude that:

\[ \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma = 18 \sin(\alpha + \beta + \gamma) \]

### Final Result

Thus, the value of \( \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma \) is:

\[ \boxed{0} \]