Question

If $\cos \alpha + 2 \cos \beta + 3 \cos \gamma = 0,\sin \alpha + 2 \sin \beta + 3 \sin \gamma = 0 $ and $ \alpha + \beta + \gamma = \pi, $ then $ \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma$ =

• A. -18
• B. 0
• C. 3
• D. 9

Answer 1

Answer: (B)

Let $a=cos\, \alpha \,+ \,i \,sin\, \alpha$
$b = cos \,\beta \,+\, i \,sin \,\beta$
and $c = cos\, \gamma\, + \,i \,sin \,\gamma$
Then, $a + 2b + 3c = \left( cos\, \alpha \,+ \,2 \,cos \,\beta + 3 \,cos\, \gamma\right) \,+\, i \left(sin \,\alpha\,+ \,2\,sin \,\beta \,+ \,3 \,sin \,\gamma\right)= 0$
$\Rightarrow a^{3} + 8b^{3} + 27c^{3} = 18abc$
$(\because if \,x + y + z = 0 $
$\Rightarrow x^{3} + y^{3} + z^{3} = 3xyz)$
$\Rightarrow cos \,3\,\alpha + 8 \,cos \,3\,\beta\, + \,27 \,cos \,3\gamma$
$= 18 cos \left(\alpha\, +\,\beta \,+ \,\gamma\right)$
and $sin \,3\,\alpha \,+ 8 \,sin\, 3\,\beta \,+ \,27\, sin\, 3\gamma$
$= 18 \,sin \left(\alpha\, +\, \beta + \gamma\right)$
$= 18 \,sin \,\pi = 0$