Question:
If \(\cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0\) then the value of \(\cos 3\alpha + \cos 3\beta + \cos 3\gamma\) is
If \(\cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0\) then the value of \(\cos 3\alpha + \cos 3\beta + \cos 3\gamma\) is
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Use complex numbers: \(z_1+z_2+z_3=0 \Rightarrow z_1^3+z_2^3+z_3^3=3z_1z_2z_3\).
Updated On: Apr 23, 2026
- \(3\cos(\alpha+\beta+\gamma)\)
- \(3\sin(\alpha+\beta+\gamma)\)
- \(3\cos(\alpha+\beta+\gamma)\)
- 0
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The Correct Option is C
Solution and Explanation
Step 1: Formula / Definition}
\[ z_1 = e^{i\alpha}, z_2 = e^{i\beta}, z_3 = e^{i\gamma} \]
Step 2: Calculation / Simplification}
\(\sum \cos \alpha + i\sum \sin \alpha = z_1 + z_2 + z_3 = 0\)
For \(z_1+z_2+z_3=0\), \(z_1^3 + z_2^3 + z_3^3 = 3z_1z_2z_3\)
\((\cos 3\alpha + i\sin 3\alpha) + (\cos 3\beta + i\sin 3\beta) + (\cos 3\gamma + i\sin 3\gamma) = 3e^{i(\alpha+\beta+\gamma)}\)
Real part: \(\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3\cos(\alpha+\beta+\gamma)\)
Step 3: Final Answer
\[ 3\cos(\alpha+\beta+\gamma) \]
\[ z_1 = e^{i\alpha}, z_2 = e^{i\beta}, z_3 = e^{i\gamma} \]
Step 2: Calculation / Simplification}
\(\sum \cos \alpha + i\sum \sin \alpha = z_1 + z_2 + z_3 = 0\)
For \(z_1+z_2+z_3=0\), \(z_1^3 + z_2^3 + z_3^3 = 3z_1z_2z_3\)
\((\cos 3\alpha + i\sin 3\alpha) + (\cos 3\beta + i\sin 3\beta) + (\cos 3\gamma + i\sin 3\gamma) = 3e^{i(\alpha+\beta+\gamma)}\)
Real part: \(\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3\cos(\alpha+\beta+\gamma)\)
Step 3: Final Answer
\[ 3\cos(\alpha+\beta+\gamma) \]
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