Question

Find the maximum and minimum value of $\sin x\sin \left({60}^0-x\right)\sin \left({60}^0+x\right).$

Answer 1

Consider

$y=\sin x\sin ({60}^0-x)\sin ({60}^0+x)$

$\Rightarrow$ $y=\sin x[{\sin }^2{60}^0-{\sin }^{2}x]$ since $\sin (A-B)\sin (A+B)={\sin }^{2}A-{\sin }^{2}B$

$\Rightarrow$

$\Rightarrow y=\sin x[{\left(\frac{\sqrt{3}}{2}\right)}^2-{\sin }^{2}x]$

$\Rightarrow y=\sin x[\frac{3}{4}-{\sin }^{2}x]$

$\Rightarrow y=\frac{1}{4}\sin x[3-4{\sin }^{2}x]$

$\Rightarrow y=\frac{1}{4}[3\sin x-4{\sin }^{3}x]$

$\Rightarrow y=\frac{1}{4}\sin 3x$ Since $[\sin 3\theta =3\sin \theta -4{\sin }^3\theta ]$

For maximum value of $\sin 3x=1$

So, maximum value of $y=\frac{1}{4}\times 1=\frac{1}{4}$

For minimum value of $\sin 3x=-1$

So, minimum value of $y=\frac{1}{4}\times (-1)=-\frac{1}{4}$