Question

If $\cos \alpha + 2\cos \beta + 3\cos \lambda = 0,{\text{ }}\sin \alpha + 2\sin \beta + 3\sin \lambda = 0{\text{ and }}\alpha + \beta + \lambda = \pi ,$ then
$
  \sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda = \\
  {\text{a}}.{\text{ - 18}} \\
  {\text{b}}{\text{. 0}} \\
  {\text{c}}{\text{. 3}} \\
  {\text{d}}{\text{. 9}} \\
$

Answer 1

Hint: Assume ${z_1} = \cos \alpha + i\sin \alpha ,{\text{ }}{{\text{z}}_2} = \cos \beta + i\sin \beta ,{\text{ }}{{\text{z}}_3} = \cos \lambda + i\sin \lambda $. Manipulate these terms to obtain the required expression, also use Euler’s Theorem that is write complex numbers in exponential terms.

Complete step-by-step answer:
Let,
  $
  {z_1} = \cos \alpha + i\sin \alpha ,{\text{ }}{{\text{z}}_2} = \cos \beta + i\sin \beta ,{\text{ }}{{\text{z}}_3} = \cos \lambda + i\sin \lambda \\
  {z_1} = \cos \alpha + i\sin \alpha .........\left( 1 \right) \\
  {\text{2}}{{\text{z}}_2} = 2\left( {\cos \beta + i\sin \beta } \right)............\left( 2 \right) \\
  {\text{3}}{{\text{z}}_3} = 3\left( {\cos \lambda + i\sin \lambda } \right)...............\left( 3 \right) \\
$
Where $z$ is a complex number
Add these three equations
$
  {z_1} + 2{{\text{z}}_2} + 3{{\text{z}}_3} = \cos \alpha + i\sin \alpha + 2\cos \beta + 2i\sin \beta + 3\cos \lambda + 3i\sin \lambda \\
  {\text{ }} = \left( {\cos \alpha + 2\cos \beta + 3\cos \lambda } \right) + i\left( {\sin \alpha + 2\sin \beta + 3\sin \lambda } \right) \\
$
Now it is given that $\cos \alpha + 2\cos \beta + 3\cos \lambda = 0,{\text{ }}\sin \alpha + 2\sin \beta + 3\sin \lambda = 0$
$ \Rightarrow {z_1} + 2{{\text{z}}_2} + 3{{\text{z}}_3} = 0 + i0 = 0$
Now according to Euler’s Theorem $\cos \alpha + i\sin \alpha = {e^{i\alpha }}$
$
   \Rightarrow {z_1} = {e^{i\alpha }} \\
   \Rightarrow {{\text{z}}_2} = {e^{i\beta }} \\
   \Rightarrow {{\text{z}}_3} = {e^{i\lambda }} \\
$
Now according to known fact if $a + b + c = 0$, then ${a^3} + {b^3} + {c^3} = 3abc$
$
   \Rightarrow {\left( {{z_1}} \right)^3} + {\left( {2{{\text{z}}_2}} \right)^3} + {\left( {3{{\text{z}}_3}} \right)^3} = 3\left( {{z_1}} \right)\left( {2{{\text{z}}_2}} \right)\left( {3{{\text{z}}_3}} \right) \\
   \Rightarrow {e^{3i\alpha }} + 8{e^{3i\beta }} + 27{e^{3i\lambda }} = 18{e^{i\alpha }}{e^{i\beta }}{e^{i\lambda }} \\
   \Rightarrow \cos 3\alpha + i\sin 3\alpha + 8\cos 3\beta + 8i\sin 3\beta + 27\cos 3\lambda + 27i\sin 3\lambda = 18{e^{i\alpha }}{e^{i\beta }}{e^{i\lambda }} \\
   \Rightarrow \left( {\cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda } \right) + i\left( {\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda } \right) = 18{e^{i\left( {\alpha + \beta + \lambda } \right)}} \\
$
Now it is given that $\alpha + \beta + \lambda = \pi $
$
   \Rightarrow \left( {\cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda } \right) + i\left( {\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda } \right) = 18{e^{i\left( {\alpha + \beta + \lambda } \right)}} = 18{e^{i\pi }} \\
   \Rightarrow \left( {\cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda } \right) + i\left( {\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda } \right) = 18\left( {\cos \pi + i\sin \pi } \right) \\
$
Now, as we know $\cos \pi = - 1,{\text{ }}\sin \pi = 0$
$ \Rightarrow \left( {\cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda } \right) + i\left( {\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda } \right) = 18\left( {\cos \pi + i\sin \pi } \right) = 18\left( { - 1 + 0i} \right) = - 18$Now comparing real and imaginary terms
$
  \cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda = - 18, \\
  \sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda = 0 \\
$
So, the value of $\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda = 0$
Hence, option b is correct.

Note: In such types of question always assume a complex number in the form of$\left( {z = \cos \theta + i\sin \theta } \right)$, then remember the standard known fact which is if $a + b + c = 0$, then ${a^3} + {b^3} + {c^3} = 3abc$, then simplify according to given conditions then compare real and imaginary parts we will get the required answer.