Q103 / 200 · Manipal (MET) 2013 · Mathematics
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
- A$0$
- B$\cos (\alpha+\beta+\gamma)$
- Correct answerC$3 \cos (\alpha+\beta+\gamma)$
- D$3 \sin (\alpha+\beta+\gamma)$
Solution
$\begin{aligned} & \text { Let } a=\cos \alpha+i \sin \alpha \\ & \qquad \begin{aligned} b & =\cos \beta+i \sin \beta \\ c & =\cos \gamma+i \sin \gamma\end{aligned}\end{aligned}$
then; $a+b+c=(\cos \alpha+\cos \beta+\cos \gamma)+$
$i(\sin \alpha+\sin \beta+\sin \gamma)$
$=0+i 0=0$
$\Rightarrow a^3+b^3+c^3=3 a b c$
$\begin{aligned} & \Rightarrow(\cos 3 \alpha+i \sin 3 \alpha)+(\cos 3 \beta \\ &+i \sin 3 \beta) \\ &+(\cos 3 \gamma+i \sin 3 \gamma)\end{aligned}$
$=3[\cos (\alpha+\beta+\gamma)+i \sin (\alpha+\beta+\gamma)]$
$\Rightarrow \cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma)$
Hence, the correct answer is option: 'C'.