Question

find maximum and minimum value of sinxsin (${60}^{\circ }$-x) sin(${60}^{\circ }$+x).

Answer 1

$lety=\sin x\sin \left(60-x\right)\sin \left(60+x\right)$

$y=\sin x\left({\sin }^{2}60-{\sin }^{2}x\right)$

$\sin x\left({\left(\frac{\sqrt{3}}{2}\right)}^2-{\sin }^{2}x\right)$

$\sin x\left(\frac{3}{4}-{\sin }^{2}x\right)$

$\frac{1}{4}\left(3\sin x-4{\sin }^{3}x\right)$

$\frac{1}{4}\sin 3x$

$Max.valueof\sin 3x=1,Maxy=\frac{1}{4}$

$NowMin.valueof\sin 3x=-1,$

$Miny=-\frac{1}{4}.$