Question

${\sin ^3}x - {\sin ^3}\left( {{{240}^0} - x} \right) + {\sin ^3}\left( {{{240}^0} + x} \right) \in \left[ { - k,k} \right] \Rightarrow k = \\\$
${\text{A}}{\text{. 1}} \\\ {\text{B}}{\text{. }}\dfrac{1}{4} \\\ {\text{C}}{\text{. }}\dfrac{3}{4} \\\ {\text{D}}{\text{. }}\dfrac{5}{4} \\\$

Answer 1

As we can see angle 240 is big so first of all make it small by writing $\left( {{{240}^0} = {{180}^0} + {{60}^0}} \right)$. Now further calculation will be a little bit easier. In question cube of sine is present so we will proceed through the formulae $\left( {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)} \right)$. And we should also remember sum to product identity
$\left( {\sin \left( {A - B} \right) - \sin \left( {A + B} \right) = - 2\cos A\sin B} \right)$ which will help in solving further.

Complete step-by-step answer:
We have
${\sin ^3}x - {\sin ^3}\left( {{{240}^0} - x} \right) + {\sin ^3}\left( {{{240}^0} + x} \right)$
We can write $\left( {{{240}^0} = {{180}^0} + {{60}^0}} \right)$
$\Rightarrow {\sin ^3}x - {\sin ^3}\left( {{{180}^0} + {{60}^0} - x} \right) + {\sin ^3}\left( {{{180}^0} + {{60}^0} + x} \right)$
We know $\left( {\sin \left( {{{180}^0} + x} \right) = - \sin x,\sin \left( {{{180}^0} - x} \right) = \sin x} \right)$
$\Rightarrow {\sin ^3}x + {\sin ^3}\left( {{{60}^0} - x} \right) - {\sin ^3}\left( {{{60}^0} + x} \right)$
We know $\left( {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)} \right)$
$\Rightarrow {\sin ^3}x + \left( {\sin \left( {{{60}^0} - x} \right) - \sin \left( {{{60}^0} + x} \right)} \right)\left( {{{\sin }^2}\left( {{{60}^0} - x} \right) + {{\sin }^2}\left( {{{60}^0} + x} \right) + \sin \left( {{{60}^0} - x} \right)\sin \left( {{{60}^0} + x} \right)} \right)$ $\ldots \ldots \left( 1 \right)$
We know $\left( {\sin \left( {A - B} \right) - \sin \left( {A + B} \right) = - 2\cos A\sin B} \right)$
This sum to product identity is very important and should always be remembered. In this identity when we put A = ${60^0}$ and B = x , then we get the following expression.
$\left( {\therefore \sin \left( {{{60}^0} - x} \right) - \sin \left( {{{60}^0} + x} \right) = - 2\cos {{60}^0}\sin x} \right)$
$\because {\sin ^2}\theta + {\sin ^2}\left( {{{60}^0} + \theta } \right) + {\sin ^2}\left( {{{60}^0} - \theta } \right) = \dfrac{3}{2} \\\ \therefore {\sin ^2}\left( {{{60}^0} + \theta } \right) + {\sin ^2}\left( {{{60}^0} - \theta } \right) = \dfrac{3}{2} - {\sin ^2}\theta \\\$
Using all these results in equation $\left( 1 \right)$ we get
$\Rightarrow {\sin ^3}x + \left( { - 2\cos {{60}^0}.\sin x} \right)\left( {\dfrac{3}{2} - {{\sin }^2}x + \sin \left( {{{60}^0} - x} \right)\sin \left( {{{60}^0} + x} \right)} \right)$
$\sin \theta .\sin \left( {{{60}^0} - \theta } \right)\sin \left( {{{60}^0} + \theta } \right) = \dfrac{1}{4}\sin 3\theta \\\ \sin \left( {{{60}^0} - \theta } \right)\sin \left( {{{60}^0} + \theta } \right) = \dfrac{{1\left( {\sin 3\theta } \right)}}{{4.\sin \theta }} = \dfrac{{3\sin \theta - 4{{\sin }^3}\theta }}{{4\sin \theta }} = \dfrac{3}{4} - {\sin ^2}\theta \\\$
$\Rightarrow {\sin ^3}x + \left( { - \sin x} \right)\left( {\dfrac{3}{2} - {{\sin }^2}x + \dfrac{{3 - 4{{\sin }^2}x}}{4}} \right)$
$\Rightarrow {\sin ^3}x - \dfrac{3}{2}\sin x + {\sin ^3}x - \dfrac{3}{4}\sin x + {\sin ^3}x$
$\Rightarrow 3{\sin ^3}x - \dfrac{9}{4}\sin x = \dfrac{{12{{\sin }^3}x - 9\sin x}}{4}$
$\Rightarrow \dfrac{3}{4}\left( {4{{\sin }^3}x - 3\sin x} \right) = \dfrac{3}{4}\left( { - \sin 3x} \right)$ $\left( {\because \sin 3x = 3\sin x - 4{{\sin }^3}x} \right)$
We know range of $\sin 3x$is $\left[ { - 1,1} \right]$
So, range of $\dfrac{3}{4}\left( { - \sin 3x} \right) = \left[ {\dfrac{{ - 3}}{4},\dfrac{3}{4}} \right]$
So $k = \dfrac{3}{4}$.hence option ${\text{C}}$ is the correct option.

Note: Whenever you get this type of question the key concept of solving is you have to use trigonometric results to shorten the complex equation and remember all trigonometric formulae to use whenever required in between solutions. These are tough questions so if you want to solve you should remember tough formula like$\left( {{{\sin }^2}\theta + {{\sin }^2}\left( {{{60}^0} + \theta } \right) + {{\sin }^2}\left( {{{60}^0} - \theta } \right) = \dfrac{3}{2}} \right)$. In this question we have used sum to product identity $\left( {\sin \left( {A - B} \right) - \sin \left( {A + B} \right) = - 2\cos A\sin B} \right)$ this can be easily proved by writing the formula of sin (A+B) and sin (A-B).