Question

The value of \[{{\sin }^{3}}({{10}^{\circ }})+{{\sin }^{3}}({{50}^{\circ }})-{{\sin }^{3}}({{70}^{\circ }})\] is
A. 0
B. \[\dfrac{5}{8}\]
C. \[\dfrac{-3}{8}\]
D. \[\dfrac{-5}{8}\]

Answer 1

Hint: The most important formulae that will be used in solving this question are as follows:
\[\Rightarrow {{\sin }^{3}}\theta =\dfrac{1}{4}\left( 3\sin \theta -\sin \left( 3\theta \right) \right)\]
\[\Rightarrow \sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)\]
In this question, we will first write the value of \[{{\sin }^{3}}x\] with the given values of x , using the relation that is given above. Then we will further use some basic relations and values of trigonometric functions to get the expression simplified. Then, we will use the formula to find the difference of two sin functions as given above and hence, we will get the final answer.

Complete step-by-step answer:
In this question, we are asked to find the value of the given trigonometric expression which is as follows
\[{{\sin }^{3}}({{10}^{\circ }})+{{\sin }^{3}}({{50}^{\circ }})-{{\sin }^{3}}({{70}^{\circ }})\]
Now, we will use the given formula as follows
\[\Rightarrow {{\sin }^{3}}\theta =\dfrac{1}{4}\left( 3\sin \theta -\sin \left( 3\theta \right) \right)\]
Now, we can write the given expression as follows using the above relation
 \[\begin{align}
  & \Rightarrow {{\sin }^{3}}({{10}^{\circ }})+{{\sin }^{3}}({{50}^{\circ }})-{{\sin }^{3}}({{70}^{\circ }}) \\
 & \Rightarrow \dfrac{1}{4}\left( -\sin \left( 3\times {{10}^{\circ }} \right)+3\sin {{10}^{\circ }} \right)+\dfrac{1}{4}\left( -\sin \left( 3\times {{50}^{\circ }} \right)+3\sin {{50}^{\circ }} \right)-\dfrac{1}{4}\left( -\sin \left( 3\times {{70}^{\circ }} \right)+3\sin {{70}^{\circ }} \right) \\
 & \Rightarrow \dfrac{1}{4}\left( -\sin \left( 3\times {{10}^{\circ }} \right)+3\sin {{10}^{\circ }} \right)+\dfrac{1}{4}\left( -\sin \left( 3\times {{50}^{\circ }} \right)+3\sin {{50}^{\circ }} \right)-\dfrac{1}{4}\left( -\sin \left( 3\times {{70}^{\circ }} \right)+3\sin {{70}^{\circ }} \right) \\
 & \Rightarrow \dfrac{1}{4}\left( -\sin \left( {{30}^{\circ }} \right)-\sin \left( {{150}^{\circ }} \right)+\sin \left( {{210}^{\circ }} \right) \right)+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+\left( \sin {{50}^{\circ }} \right)-\left( \sin {{70}^{\circ }} \right) \right) \\
\end{align}\]
Now, as we know that
\[\begin{align}
  & \Rightarrow \sin {{30}^{\circ }}=\dfrac{1}{2} \\
 & \Rightarrow \sin {{150}^{\circ }}=\sin \left( {{180}^{\circ }}-{{30}^{\circ }} \right)=\dfrac{1}{2} \\
 & \left[ \sin \left( {{180}^{\circ }}-x \right)=\sin x \right] \\
 & \Rightarrow \sin {{210}^{\circ }}=\sin \left( {{180}^{\circ }}+{{30}^{\circ }} \right)=\dfrac{-1}{2} \\
 & \left[ \sin \left( {{180}^{\circ }}+x \right)=-\sin x \right] \\
\end{align}\]
Hence, we can write the above expression as follows
 \[\begin{align}
  & \Rightarrow \dfrac{1}{4}\left( -\sin \left( {{30}^{\circ }} \right)-\sin \left( {{150}^{\circ }} \right)+\sin \left( {{210}^{\circ }} \right) \right)+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+\left( \sin {{50}^{\circ }} \right)-\left( \sin {{70}^{\circ }} \right) \right) \\
 & \Rightarrow \dfrac{1}{4}\left( -\dfrac{1}{2}-\dfrac{1}{2}+\left( \dfrac{-1}{2} \right) \right)+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+\left( \sin {{50}^{\circ }} \right)-\left( \sin {{70}^{\circ }} \right) \right) \\
 & \Rightarrow \dfrac{-1}{4}\left( \dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2} \right)+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+\left( \sin {{50}^{\circ }} \right)-\left( \sin {{70}^{\circ }} \right) \right) \\
 & \Rightarrow \dfrac{-1}{4}\left( \dfrac{3}{2} \right)+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+\left( \sin {{50}^{\circ }} \right)-\left( \sin {{70}^{\circ }} \right) \right) \\
 & \Rightarrow \dfrac{-3}{8}+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+\left( \sin {{50}^{\circ }} \right)-\left( \sin {{70}^{\circ }} \right) \right) \\
\end{align}\]
Now, we will use the formula that is given in the hint for subtracting two sin functions and we can write as follows
\[\begin{align}
  & \Rightarrow \dfrac{-3}{8}+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+2\left( \sin \left( \dfrac{{{50}^{\circ }}+{{70}^{\circ }}}{2} \right) \right)\left( \sin \left( \dfrac{{{50}^{\circ }}-{{70}^{\circ }}}{2} \right) \right) \right) \\
 & \left[ \sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right) \right] \\
 & \Rightarrow \dfrac{-3}{8}+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+2\left( \cos \left( \dfrac{{{120}^{\circ }}}{2} \right) \right)\left( \sin \left( \dfrac{-{{20}^{\circ }}}{2} \right) \right) \right) \\
 & \Rightarrow \dfrac{-3}{8}+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+2\left( \cos \left( {{60}^{\circ }} \right) \right)\left( \sin \left( -{{10}^{\circ }} \right) \right) \right) \\
 & \Rightarrow \dfrac{-3}{8}+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+2\left( \cos \left( {{60}^{\circ }} \right) \right)\left( \sin \left( -{{10}^{\circ }} \right) \right) \right) \\
\end{align}\]
Now, we know that the value of \[\cos {{60}^{\circ }}\] is \[\dfrac{1}{2}\] , hence, we can write it as follows
 \[\begin{align}
  & \Rightarrow \dfrac{-3}{8}+\dfrac{3}{4}\left( \sin {{10}^{\circ }}+2\times \dfrac{\sqrt{3}}{2}\left( \sin \left( -{{10}^{\circ }} \right) \right) \right) \\
 & \Rightarrow \dfrac{-3}{8}+\dfrac{3}{4}\left( \sin {{10}^{\circ }}-\left( \sin \left( {{10}^{\circ }} \right) \right) \right) \\
 & \left[ \sin \left( -x \right)=-\sin x \right] \\
 & \Rightarrow \dfrac{-3}{8}+\dfrac{3}{4}\times 0 \\
 & \Rightarrow \dfrac{-3}{8} \\
\end{align}\]


NOTE:-
For solving this question correctly, the students must know about the trigonometric functions like some of the trigonometric relations such as the following
\[\begin{align}
  & \left[ \sin \left( {{180}^{\circ }}-x \right)=\sin x \right] \\
 & \left[ \sin \left( {{180}^{\circ }}+x \right)=-\sin x \right] \\
\end{align}\]
Without knowing these values, one can never get to the correct answer.