Question

Check the following result:
\[{{\sin }^{3}}A+{{\sin }^{3}}\left( \dfrac{2\pi }{3}-A \right)+{{\sin }^{3}}\left( \dfrac{4\pi }{3}-A \right)=-\dfrac{4}{3}\sin 3A\]

Answer 1

Hint: We solve this problem by using the composite angles formulas.
We have the formula of sine ratio as
\[\sin 3x=3\sin x-4{{\sin }^{3}}x\]
By using the above formula we take the LHS of the given equation and prove that LHS is equal to RHS.
We also have the composite angle formula for sine ratio as
\[\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y\]

Complete step by step answer:
We are asked to prove the result
\[{{\sin }^{3}}A+{{\sin }^{3}}\left( \dfrac{2\pi }{3}-A \right)+{{\sin }^{3}}\left( \dfrac{4\pi }{3}-A \right)=-\dfrac{4}{3}\sin 3A\]
Let us assume that LHS of the given equation as
\[\Rightarrow LHS={{\sin }^{3}}A+{{\sin }^{3}}\left( \dfrac{2\pi }{3}-A \right)+{{\sin }^{3}}\left( \dfrac{4\pi }{3}-A \right)\]
We know that the formula of sine ratio as
\[\sin 3x=3\sin x-4{{\sin }^{3}}x\]
Now, let us find the value of cube of sine ratio in terms of sine of three times the angle that is
\[\begin{align}
  & \Rightarrow \sin 3x=3\sin x-4{{\sin }^{3}}x \\
 & \Rightarrow 4{{\sin }^{3}}x=3\sin x-\sin 3x \\
 & \Rightarrow {{\sin }^{3}}x=\dfrac{3\sin x-\sin 3x}{4} \\
\end{align}\]
By using the above formula of cube of sine ratio to each term in the LHS then we get

\[\begin{align}
  & \Rightarrow LHS=\left( \dfrac{3\sin A-\sin 3A}{4} \right)+\left( \dfrac{3\sin \left( \dfrac{2\pi }{3}-A \right)-\sin 3\left( \dfrac{2\pi }{3}-A \right)}{4} \right)+\left( \dfrac{3\sin \left( \dfrac{4\pi }{3}-A \right)-\sin 3\left( \dfrac{4\pi }{3}-A \right)}{4} \right) \\
 & \Rightarrow LHS=\left( \dfrac{3\sin A-\sin 3A}{4} \right)+\left( \dfrac{3\sin \left( \dfrac{2\pi }{3}-A \right)-\sin \left( 2\pi -3A \right)}{4} \right)+\left( \dfrac{3\sin \left( \dfrac{4\pi }{3}-A \right)-\sin \left( 4\pi -3A \right)}{4} \right) \\
\end{align}\]We know that the standard result of change of ratios that is
\[\sin \left( n\pi -x \right)=\left\{ \begin{align}
  & \sin x,\forall n\in odd \\
 & -\sin x,\forall n\in even \\
\end{align} \right.\]
By using this relation to above equation we get
\[\begin{align}
  & \Rightarrow LHS=\left( \dfrac{3\sin A-\sin 3A}{4} \right)+\left( \dfrac{3\sin \left( \dfrac{2\pi }{3}-A \right)+\sin 3A}{4} \right)+\left( \dfrac{3\sin \left( \dfrac{4\pi }{3}-A \right)+\sin 3A}{4} \right) \\
 & \Rightarrow LHS=\dfrac{3}{4}\left( \sin A+\sin \left( \dfrac{2\pi }{3}-A \right)+\sin \left( \dfrac{4\pi }{3}-A \right) \right)+\left( -\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4} \right)\sin 3A \\
 & \Rightarrow LHS=\dfrac{3}{4}\left[ \sin A+\sin \left( \dfrac{2\pi }{3}-A \right)+\sin \left( \dfrac{4\pi }{3}-A \right) \right]+\dfrac{1}{4}\sin \left( 3A \right) \\
\end{align}\]
We know that the formula of composite angles that is
\[\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y\]
By using these formulas to above equation of LHS we get
\[\Rightarrow LHS=\dfrac{3}{4}\left[ \sin A+\left( \sin \dfrac{2\pi }{3}\cos A-\cos \dfrac{2\pi }{3}\sin A \right)+\left( \sin \dfrac{4\pi }{3}\cos A-\cos \dfrac{4\pi }{3}\sin A \right) \right]+\dfrac{1}{4}\sin 3A\]
We also know that the values of some standard angles of sine and cosine ratios as
\[\begin{align}
  & \Rightarrow \sin \dfrac{2\pi }{3}=\dfrac{\sqrt{3}}{2},\sin \dfrac{4\pi }{3}=-\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \cos \dfrac{2\pi }{3}=\cos \dfrac{4\pi }{3}=-\dfrac{1}{2} \\
\end{align}\]
By using these formulas to above equation of LHS we get
\[\begin{align}
  & \Rightarrow LHS=\dfrac{3}{4}\left[ \sin A+\left( \sin \dfrac{2\pi }{3}\cos A-\cos \dfrac{2\pi }{3}\sin A \right)+\left( \sin \dfrac{4\pi }{3}\cos A-\cos \dfrac{4\pi }{3}\sin A \right) \right]+\dfrac{1}{4}\sin 3A \\
 & \Rightarrow LHS=\dfrac{3}{4}\left[ \sin A+\sin A \right]+\dfrac{1}{4}\left( \sin 3A \right) \\
 & \Rightarrow LHS=\dfrac{3}{2}\sin A+\dfrac{1}{4}\left( \sin 3A \right) \\
\end{align}\]
Here we can see that values of LHS and RHS are not equal.
Therefore we can conclude that the required result is not correct that is
\[\therefore {{\sin }^{3}}A+{{\sin }^{3}}\left( \dfrac{2\pi }{3}-A \right)+{{\sin }^{3}}\left( \dfrac{4\pi }{3}-A \right)\ne -\dfrac{4}{3}\sin 3A\]

Note:We can prove this result by taking the example of the angle.
We are asked to prove
\[{{\sin }^{3}}A+{{\sin }^{3}}\left( \dfrac{2\pi }{3}-A \right)+{{\sin }^{3}}\left( \dfrac{4\pi }{3}-A \right)=-\dfrac{4}{3}\sin 3A\]
Let us assume some angle \[A=\dfrac{\pi }{3}\]
By substituting this angle in LHS we get
\[\begin{align}
  & \Rightarrow LHS={{\sin }^{3}}\left( \dfrac{\pi }{3} \right)+{{\sin }^{3}}\left( \dfrac{2\pi }{3}-\dfrac{\pi }{3} \right)+{{\sin }^{3}}\left( \dfrac{4\pi }{3}-\dfrac{\pi }{3} \right) \\
 & \Rightarrow LHS={{\sin }^{3}}\left( \dfrac{\pi }{3} \right)+{{\sin }^{3}}\left( \dfrac{\pi }{3} \right)+{{\sin }^{3}}\left( \pi \right) \\
\end{align}\]
We know that values of angles that are
\[\begin{align}
  & \sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2} \\
 & \sin n\pi =0 \\
\end{align}\]
By using the above values in the equation of LHS we get
\[\begin{align}
  & \Rightarrow LHS={{\left( \dfrac{\sqrt{3}}{2} \right)}^{3}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{3}}+0 \\
 & \Rightarrow LHS=\dfrac{3\sqrt{3}}{2} \\
\end{align}\]
Similarly by substituting the value \[A=\dfrac{\pi }{3}\] in RHS we get
\[\begin{align}
  & \Rightarrow RHS=-\dfrac{4}{3}\sin \left( 3\times \dfrac{\pi }{3} \right) \\
 & \Rightarrow RHS=-\dfrac{4}{3}\sin \left( \pi \right)=0 \\
\end{align}\]
Here we can see that values of LHS and RHS are not equal.
Therefore we can conclude that the required result is not correct
\[\therefore {{\sin }^{3}}A+{{\sin }^{3}}\left( \dfrac{2\pi }{3}-A \right)+{{\sin }^{3}}\left( \dfrac{4\pi }{3}-A \right)=-\dfrac{4}{3}\sin 3A\]
This method is followed only in case of objective type questions not for subjective.