Question

The value of ${\sin ^3}{10^0} + {\sin ^3}{50^0} - {\sin ^3}{70^0}$ is equal to:
$A.\,\,\dfrac{1}{8} \\\ B.\,\,\dfrac{{ - 3}}{8} \\\ C.\,\,\dfrac{{ - 1}}{8} \\\ D.\,\,None\,\,of\,\,these \\\$

Answer 1

Hint : To find value we first show that value of $\sin {10^0} + \sin {50^0} - \sin {70^0}$ is zero, then writing ${\sin ^3}{10^0} + {\sin ^3}{50^0} - {\sin ^3}{70^0}$ as $3\sin {10^0}\sin {50^0}\left( { - \sin {{70}^0}} \right)$ by using algebraic identity and finally solving it by using identity of trigonometry to get required value of the problem.
$\left( {{A^3} + {B^3} + {C^3} - 3ABC} \right) = \left( {A + B + C} \right)\left( {{A^2} + {B^2} + {C^2} - AB - BC - CA} \right)$ , $\sin 3A = 3\sin A - 4{\sin ^3}A$
And $\sin C - \sin D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$ , $\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B$

Complete step-by-step answer :
We have given,
${\sin ^3}{10^0} + {\sin ^3}{50^0} - {\sin ^3}{70^0}$
Or we can write it as:
${\left( {\sin {{10}^0}} \right)^3} + {\left( {\sin {{50}^0}} \right)^3} + {\left( { - \sin {{70}^0}} \right)^3}$
Or
{\left( {\sin {{10}^0}} \right)^3} + {\left( {\sin {{50}^0}} \right)^3} + {\left\\{ {\sin \left( { - {{70}^0}} \right)} \right\\}^3}
Also, from identity we have:
${A^3} + {B^3} + {C^3} = 3ABC,\,\,if\,\,A + B + C = 0$
Therefore, to get solution of the problem, we just show that $\sin {10^0} + \sin {50^0} + \sin ( - {70^0}) = 0$
Now, solving $\sin {10^0} + \sin {50^0} + \sin ( - {70^0})$
We can write it as:
\sin {10^0} + \sin {50^0} - \sin {70^0} \left\\{ {\sin ( - \theta ) = - \sin \theta } \right\\}
Applying identity to simplify
2\sin \left\\{ {\dfrac{{10 + 50}}{2}} \right\\}\cos \left\\{ {\dfrac{{10 - 50}}{2}} \right\\} - \sin {70^0} \\\ \Rightarrow 2\sin {30^0} \cos \left( { - {{20}^0}} \right) - \sin {70^0} \left\\{ {\cos ( - \theta ) = \cos \theta } \right\\} \\\ \Rightarrow 2\times\dfrac{1}{2}\cos {20^0} - \sin {70^0} \\\
$\Rightarrow \cos {20^0} - \sin {70^0}$
\Rightarrow \sin \left( 90^0 - 20^0 \right) - \sin {70^0} \left\\{ {\cos \theta = \sin \left( {{{90}^0}\theta } \right)} \right\\}
$\Rightarrow \sin {70^0} - \sin {70^0} = 0$
Therefore, we have
${\sin ^3}{10^0} + {\sin ^3}{50^0} - {\sin ^3}{70^0} = 3.\left( {\sin {{10}^0}} \right)\left( {\sin {{50}^0}} \right)\left( { - \sin {{70}^0}} \right)$
Now, simplifying the right hand side of the above equation. We have
\- 3\sin {10^0}\sin {50^0}\sin {70^0} \\\ \Rightarrow - 3\sin {10^0}\sin \left( {60 - 10} \right)\sin \left( {60 + 10} \right) \\\ \Rightarrow - 3\sin {10^0}\left\\{ {{{\sin }^2}{{60}^0} - {{\sin }^2}10} \right\\} \\\ \Rightarrow - 3\sin {10^0}\left\\{ {{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2} - {{\sin }^2}{{10}^0}} \right\\} \\\ \Rightarrow - 3\sin {10^0}\left\\{ {\dfrac{3}{4} - {{\sin }^2}{{10}^0}} \right\\} \\\ \Rightarrow - 3\sin {10^0}\left( {\dfrac{{3 - 4{{\sin }^2}10}}{4}} \right) \\\ \Rightarrow - 3\left( {\dfrac{{3\sin {{10}^0} - 4{{\sin }^3}10}}{4}} \right) \\\ \Rightarrow \dfrac{{ - 3}}{4} \times \sin 3\left( {{{10}^0}} \right) \\\ \Rightarrow \dfrac{{ - 3}}{4}\sin {30^0} \\\ \Rightarrow - \dfrac{3}{4} \times \dfrac{1}{2} \\\ \Rightarrow - \dfrac{3}{8} \;
Hence, from above we have:
${\sin ^3}{10^0} + {\sin ^3}{50^0} - {\sin ^3}{70^0} = - \dfrac{3}{8}$
Therefore, required value of ${\sin ^3}{10^0} + {\sin ^3}{50^0} - {\sin ^3}{70^0}\,is - \dfrac{3}{8}$
So, the correct answer is “Option B”.

Note : Solution of the given problems can also be found in other ways. In this we use a different trigonometric identity to find its value. In this we substitute value of ${\sin ^3}A,\,\,as\,\,\,\dfrac{{3\sin x - {{\sin }^3}x}}{4}$ on each term and then writing value of standard angle and simplify rest of the term by using different trigonometric formulas to find required solution.