Question

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

Answer 1

Let us consider the problem:

Let $y=\sin x.\sin \left(60-x\right).\sin \left(60-x\right)$

$\Rightarrow y=\sin x\left[{\sin }^2{60}^{\circ }-{\sin }^{2}x\right]$ $(\therefore \sin \left(A-B\right)\sin \left(A+B\right)={\sin }^{2}A-{\sin }^{2}B)$

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

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

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

$\Rightarrow y=\frac{1}{4}\left[3\sin x-\left(3\sin x-\sin 3x\right)\right]$ $(\therefore \sin 3x=3\sin x-4{\sin }^{3}x)$

$\Rightarrow$$y=\frac{1}{4}\sin 3x$

Maximum value of $sin3x=1$

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

and minimum value of $sin3x=-1$

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