نکته $:\,\operatorname{Sin}2x=2\operatorname{Sin}x\operatorname{Cos}x\,\,\,\,\,\,\,,\,\,\,\,\,\,\,{{\operatorname{Cos}}^{2}}x-{{\operatorname{Sin}}^{2}}x=\operatorname{Cos}2x$
ابتدا عبارت $A$ را ساده میکنیم:
$A=\operatorname{Sin}x{{\operatorname{Cos}}^{3}}x-\operatorname{Cos}{{\operatorname{Sin}}^{3}}x=\operatorname{Sin}x\operatorname{Cos}x({{\operatorname{Cos}}^{2}}x-{{\operatorname{Sin}}^{2}}x)=\frac{1}{2}\operatorname{Sin}2x\times \operatorname{Cos}2x=\frac{1}{2}\times \frac{1}{2}\operatorname{Sin}4x=\frac{1}{4}\operatorname{Sin}4x$
حال با جایگذاری $x=\frac{\pi }{16}$ داریم:
$A=\frac{1}{4}\operatorname{Sin}(4\times \frac{\pi }{16})=\frac{1}{4}\operatorname{Sin}\frac{\pi }{4}=\frac{1}{4}\times \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{8}$