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If $\cos\theta+2\cos\alpha+3\cos\beta=\sin\theta+2\sin\alpha+3\sin\beta=0$, then $\sin\;3\theta+8\sin\;3\alpha+27\sin\;3\beta=?$

(A) $9\;\sin(\theta+\alpha+\beta)$ (B) $18\;\sin(\theta+\alpha+\beta)$ (C) $18\;\cos (\theta+\alpha+\beta)$ (D) $9\;\cos (\theta+\alpha+\beta)$
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Toolbox:
  • If $x+y+z=0$, then $x^3+y^3+z^3=3xyz$
  • $(\cos\theta+i \sin\theta)^n=\cos\;n\theta+i \sin\;n\theta$
Let $x=cos\theta+isin\theta,\:y=2(cos\alpha+isin\alpha)$, $z=3(cos\beta+isin\beta)$
$\Rightarrow\:x+y+z=(cos\theta+2cos\alpha+3cos\beta)+i(sin\theta+2sin\alpha+3sin\beta)=0$
$\Rightarrow\:x+y+z=0$
$\Rightarrow\:x^3+y^3+z^3=3xyz$
$\Rightarrow\:(cos\theta+isin\theta)^3+8(cos\alpha+isin\alpha)^3+27(cos\beta+isin\beta)^3$
$=3.(cos\theta+isin\theta).2(cos\alpha+isin\alpha).3(cos\beta+isin\beta)$
$\Rightarrow\:(cos\;3\theta+isin\theta)+8(cos\;3\alpha+isin\;3\alpha)+27(cos\;3\beta+isin\;3\beta)=$
$18(cos(\theta+\alpha+\beta)+isin(\theta+\alpha+\beta)$
Equating imaginary part on both the sides,
$sin\;3\theta+8sin\;3\alpha+27sin\;3\beta=18\;sin(\theta+\alpha+\beta)$
answered Aug 2, 2013 by rvidyagovindarajan_1
edited Mar 21, 2014 by balaji.thirumalai
 

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