Question

Range of ${\sin }^{3}x-{\sin }^3({360}^0-3x)+{\sin }^3({360}^0+3x)$

• A. $[-1,1]$
• B. $[-\frac{1}{4},\frac{1}{4}]$
• C. $[-\frac{3}{4},\frac{3}{4}]$
• D. None of these

Answer 1

Answer: (C)

$[-\frac{3}{4},\frac{3}{4}]$

Let $f\left(x\right)={\sin }^{3}x-\sin \left(3\right(120-x\left)\right)+\sin 3({120}^o+x)$
$={\sin }^{3}x-\sin (2\pi -3x)+\sin (2\pi +3x)$
$={\sin }^{3}x+\sin 3x+\sin 3x$
$={\sin }^{3}x+2\sin 3x$
$=2(3\sin x-4{\sin }^{3}x)+{\sin }^{3}x[\sin 3x=3\sin x-4{\sin }^{3}x]$
$f\left(x\right)=6\sin x-7{\sin }^{3}x$
$f^{\prime }\left(x\right)=6\cos x-21{\sin }^{2}x\cos x=0$
${\sin }^{2}x=\frac{6}{21}=\frac{2}{7}$
if , $\sin x=0,f\left(x\right)=0$
if , $\sin x=-1,f\left(x\right)=1$
if , $\sin x=1,f\left(x\right)=-1$
at $\sin x=-\sqrt{\frac{2}{7}}=-6\sqrt{\frac{2}{7}}+2\sqrt{\frac{2}{7}}=-4\sqrt{\frac{2}{7}}$
$\sin x=\sqrt{\frac{2}{7}}=4\sqrt{\frac{2}{7}}$
range $ϵ[-4\sqrt{\frac{2}{7}},4\sqrt{\frac{2}{7}}]$