Find maximum and minimum value of $\sin x.\sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right)$

Hint: We are having a trigonometric equation as: $\sin x.\sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right)$. The expression consists of $\sin A.\sin B.\sin C$, which can be written as: $\sin A.\left( \sin B.\sin C \right)$. Now, divide and multiply the expression by 2 to form an identity of cosine subtraction, i.e. $\cos \left( A-B \right)-\cos \left( A+B \right)=2\sin A.\sin B$. Simplify the expression by substitution the value of cosine of standard angle, if any and then using double angle relation convert the whole expression in terms of sine, i.e. $\cos 2x=1-2{{\sin }^{2}}x$. In the end, try to form an identity of $\sin 3x$ and using the domain of sine function, i.e. [-1, 1], find the minimum value or the expression.

Complete step-by-step solution:
As we are given the following expression: $\sin x.\sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right).................(1)$
We can write equation (1) as:
$\sin x.\left\{ \sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right) \right\}.............(2)$
Now, multiply and divide equation (2) by 2, we get:
$\dfrac{1}{2}\sin x\left\{ 2\sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right) \right\}..............(3)$
As we know that: $\cos \left( A-B \right)-\cos \left( A+B \right)=2\sin A.\sin B$
So, by applying cosine subtraction identity to the equation (3), we get:
$\begin{align}
  & \Rightarrow \dfrac{1}{2}\sin x\left\{ \cos \left( {{60}^{\circ }}-x-{{60}^{\circ }}-x \right)-\cos \left( {{60}^{\circ }}-x+{{60}^{\circ }}+x \right) \right\} \\
 & \Rightarrow \dfrac{1}{2}\sin x\left\{ \cos \left( -2x \right)-\cos \left( {{120}^{\circ }} \right) \right\}..............(4) \\
\end{align}$
Since, \[\cos \left( -\theta \right)=\cos \theta ;\cos \left( {{120}^{\circ }} \right)=-\dfrac{1}{2}\]
So, by putting the values in equation (4), we get:
$\begin{align}
  & \Rightarrow \dfrac{1}{2}\sin x\left\{ \cos \left( 2x \right)-\left( -\dfrac{1}{2} \right) \right\} \\
 & \Rightarrow \dfrac{1}{2}\sin x\left\{ \cos \left( 2x \right)+\dfrac{1}{2} \right\}...........(5) \\
\end{align}$
As we know that: $\cos 2x=1-2{{\sin }^{2}}x$
So, we can write equation (5) as:
$\begin{align}
  & \Rightarrow \dfrac{1}{2}\sin x\left\{ 1-2{{\sin }^{2}}x+\dfrac{1}{2} \right\} \\
 & \Rightarrow \dfrac{1}{2}\sin x\left\{ \dfrac{3}{2}-2{{\sin }^{2}}x \right\} \\
 & \Rightarrow \left\{ \dfrac{3}{4}\sin x-{{\sin }^{3}}x \right\} \\
 & \Rightarrow \dfrac{3\sin x-{4{\sin }^{3}}x}{4}...............(6) \\
\end{align}$
As we know that: $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta $
So, by applying the identity to equation (6), we get:
$\sin x.\sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right)=\dfrac{1}{4}\sin 3x$
Now, we need to find the minimum value of $\dfrac{1}{4}\sin 3x$
As we know that domain of sine of an angle is [-1, 1], so we can write that:
$-1 \le \sin \theta \le 1$
Using this result for $\sin 3x$, we can write:
$-1\le \sin 3x \le 1$
Now, divide the whole equation by 4 to get a relation for $\dfrac{1}{4}\sin 3x$, we get:
$-\dfrac{1}{4} \le \dfrac{1}{4}\sin 3x \le \dfrac{1}{4}$
So, the minimum value of $\dfrac{1}{4}\sin 3x$ is $-\dfrac{1}{4}$
Therefore, minimum value of $\sin x.\sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right)$ is $-\dfrac{1}{4}$.

Note: There is another way to solve an expression in the form of$\sin \theta .\sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)$. It is an important result or identity of trigonometry, i.e.
$\sin \theta .\sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\dfrac{1}{4}\sin 3\theta $
By using the above result for the expression: $\sin x.\sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right)$, we can write:
$\sin x.\sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right)=\dfrac{1}{4}\sin 3x$
Now, we need to find the minimum value of $\dfrac{1}{4}\sin 3x$
As we know that domain of sine of an angle is [-1, 1], so we can write that:
$-1 \le \sin \theta \le 1$
Using this result for $\sin 3x$, we can write:
$-1 \le\sin 3x \le 1$
Now, divide the whole equation by 4 to get a relation for $\dfrac{1}{4}\sin 3x$, we get:
$-\dfrac{1}{4} \le \dfrac{1}{4}\sin 3x \le \dfrac{1}{4}$
So, the minimum value of $\dfrac{1}{4}\sin 3x$ is $-\dfrac{1}{4}$
Therefore, minimum value of $\sin x.\sin \left( {{60}^{\circ }}-x \right)\sin \left( {{60}^{\circ }}+x \right)$ is $-\dfrac{1}{4}$.