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.