If ` cos A + cos B + cos C = 0, sin A + sin B + sin C = 0 and A + B + C = 180^(@) `, then the value of ` cos 3A + cos 3 B + cos 3 C is `

Options

A. 3
B. `-3`
C. `sqrt(3)`
D. 0

Answer

To solve the problem, we need to find the value of \( \cos 3A + \cos 3B + \cos 3C \) given the conditions:

1. \( \cos A + \cos B + \cos C = 0 \)
2. \( \sin A + \sin B + \sin C = 0 \)
3. \( A + B + C = 180^\circ \)

### Step-by-Step Solution:

**Step 1: Use the given equations.**

From the first two equations, we can combine them into a single complex equation:
\[ \cos A + \cos B + \cos C + i(\sin A + \sin B + \sin C) = 0 \]
This implies:
\[ e^{iA} + e^{iB} + e^{iC} = 0 \]

**Hint for Step 1:** Remember that \( e^{i\theta} = \cos \theta + i \sin \theta \).

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**Step 2: Express \( A, B, C \) in terms of complex exponentials.**

Let:
\[ z_1 = e^{iA}, \quad z_2 = e^{iB}, \quad z_3 = e^{iC} \]
Since \( z_1 + z_2 + z_3 = 0 \), we can express \( z_3 \) as:
\[ z_3 = - (z_1 + z_2) \]

**Hint for Step 2:** The sum of the roots of a polynomial can give you important relationships.

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**Step 3: Calculate \( \cos 3A + \cos 3B + \cos 3C \).**

Using the identity for cosine of a triple angle:
\[ \cos 3\theta = 4\cos^3 \theta - 3\cos \theta \]
We can express:
\[ \cos 3A + \cos 3B + \cos 3C = 4(\cos^3 A + \cos^3 B + \cos^3 C) - 3(\cos A + \cos B + \cos C) \]
Since \( \cos A + \cos B + \cos C = 0 \), we have:
\[ \cos 3A + \cos 3B + \cos 3C = 4(\cos^3 A + \cos^3 B + \cos^3 C) \]

**Hint for Step 3:** Use the identity for the sum of cubes when needed.

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**Step 4: Use the identity for the sum of cubes.**

Using the identity:
\[ \cos^3 A + \cos^3 B + \cos^3 C = \frac{1}{3}(\cos A + \cos B + \cos C)(\cos^2 A + \cos^2 B + \cos^2 C - \cos A \cos B - \cos B \cos C - \cos C \cos A) \]
Since \( \cos A + \cos B + \cos C = 0 \), we find:
\[ \cos^3 A + \cos^3 B + \cos^3 C = 0 \]

**Hint for Step 4:** Remember that if a term is multiplied by zero, the entire expression becomes zero.

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**Step 5: Conclude the result.**

Thus, substituting back, we get:
\[ \cos 3A + \cos 3B + \cos 3C = 4 \cdot 0 = 0 \]

### Final Answer:
\[ \cos 3A + \cos 3B + \cos 3C = 0 \]

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