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 find the value of \[\sin 3\alpha +8\sin 3\beta +27\sin 3\gamma \]
$\begin{align}
  & \text{a) }\text{9} \\
 & \text{b) -18} \\
 & \text{c) 0} \\
 & \text{d) 3} \\
\end{align}$

Answer 1

Hint: Now we are given with two equations one in terms of cos and one in terms of sin. So we will try to write the two equations as complex equations. To do so we multiply the equation \[\sin \alpha +2\sin \beta +3\sin \gamma =0\] with i and add the equation to \[\cos \alpha +2\cos \beta +3\cos \gamma =0\] . Now we know that $\cos \theta +i\sin \theta ={{e}^{i\theta }}$ . Hence we will substitute this and get a Complex equation. Further we can use the fact that if $a+b+c=0\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$. In this equation we again convert the equations in $\cos \theta +i\sin \theta $ form. Now we know if two complex numbers are equal then their real and imaginary parts are equal. Hence we will get the required equation in the form of $\sin (\alpha +\beta +\gamma )$ and we know $\alpha +\beta +\gamma =\pi $. Hence we can find the value of \[\sin 3\alpha +8\sin 3\beta +27\sin 3\gamma \]

Complete step by step answer:
Now let us first consider the equations.
\[\cos \alpha +2\cos \beta +3\cos \gamma =0.............(1)\]
\[\sin \alpha +2\sin \beta +3\sin \gamma =0...............(2)\]
Now multiplying equation (2) with i and then adding it to equation (1) we get
\[\cos \alpha +2\cos \beta +3\cos \gamma +i\sin \alpha +2i\sin \beta +3i\sin \gamma =0\]
Rearranging these terms we get
\[\cos \alpha +i\sin \alpha +2\cos \beta +2i\sin \beta +3\cos \gamma +3i\sin \gamma =0\]
Now writing the equation in form of $\cos \theta +i\sin \theta $ we get
\[\cos \alpha +i\sin \alpha +2(\cos \beta +i\sin \beta )+3(\cos \gamma +i\sin \gamma )=0\]
Now we know that $\cos \theta +i\sin \theta ={{e}^{i\theta }}$ hence, we can write the above equation as
${{e}^{i\alpha }}+2{{e}^{i\beta }}+3{{e}^{i\gamma }}=0$
Now let us write $a={{e}^{i\alpha }},b=2{{e}^{i\beta }},c=3{{e}^{i\gamma }}$
So we have $a+b+c=0$. Now we know that if $a+b+c=0\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$
Hence we have ${{\left( {{e}^{i\alpha }} \right)}^{3}}+{{\left( 2{{e}^{i\beta }} \right)}^{3}}+{{\left( 3{{e}^{i\gamma }} \right)}^{3}}=3\left( {{e}^{i\alpha }} \right)\left( 2{{e}^{i\beta }} \right)\left( 3{{e}^{i\gamma }} \right)$
$\left( {{e}^{i3\alpha }} \right)+\left( 8{{e}^{i3\beta }} \right)+\left( 27{{e}^{i3\gamma }} \right)=18\left( {{e}^{i\alpha +i\beta +i\gamma }} \right)$
$\left( {{e}^{i3\alpha }} \right)+\left( 8{{e}^{i3\beta }} \right)+\left( 27{{e}^{i3\gamma }} \right)=18\left( {{e}^{i(\alpha +\beta +\gamma )}} \right)$
Now again using $\cos \theta +i\sin \theta ={{e}^{i\theta }}$ we can write the equation as

\[\begin{align}
  & \cos 3\alpha +8\cos 3\beta +27\cos 3\gamma +i\sin 3\alpha +8i\sin 3\beta +27i\sin 3\gamma =18[\cos (\alpha +\beta +\gamma )+i\sin (\alpha +\beta +\gamma )] \\
 & \Rightarrow \cos 3\alpha +8\cos 3\beta +27\cos 3\gamma +i(\sin 3\alpha +8\sin 3\beta +27\sin 3\gamma )=18\cos (\alpha +\beta +\gamma )+i18\sin (\alpha +\beta +\gamma )] \\
\end{align}\]
Now we know if now we know if two complex numbers are equal then their real and imaginary parts are equal. So let us take the imaginary parts as equal and hence we get
\[\sin 3\alpha +8\sin 3\beta +27\sin 3\gamma =18\sin (\alpha +\beta +\gamma )\] .
Now we are given that $\alpha +\beta +\gamma =\pi $ this means \[\sin (\alpha +\beta +\gamma )=\sin \pi =0\]
Hence, \[\sin 3\alpha +8\sin 3\beta +27\sin 3\gamma =18\sin (\alpha +\beta +\gamma )=18\sin \pi =0\]
Hence we now get the value of \[\sin 3\alpha +8\sin 3\beta +27\sin 3\gamma \] is 0

So, the correct answer is “Option C”.

Note: Now for solving this question we have used the result if$a+b+c=0\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ This can be easily verified as we know the formula for \[{{a}^{3}}+{{b}^{3}}~+\text{ }{{c}^{3}}~=\text{ }\left( a+\text{ }b\text{ }+\text{ }c \right)\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }ab\text{ }\text{ }bc\text{ }\text{ }ca)\text{ }+\text{ }3abc\] here if we substitute $a+b+c=0$ we get ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$.