Question

${\cos }^{3}x+{\cos }^3\left({120}^0+x\right)+{\cos }^3\left({240}^0+x\right)\in$

• A. $[-1,1]$
• B. $\left[-\frac{1}{4},\frac{1}{4}\right]$
• C. $\left[-\frac{3}{4},\frac{3}{4}\right]$
• D. $[-2,2]$

Answer 1

The correct option is C $\left[-\frac{3}{4},\frac{3}{4}\right]$

$\cos A=4{\cos }^{3}A-3\cos A$

Now

${\cos }^3\left(x\right)+{\cos }^3({120}^o+x)+{\cos }^3({240}^o+x)=\frac{1}{4}(\cos 3x+3\cos x)+\frac{1}{4}[\cos ({360}^o+x)+3\cos ({120}^o+x)]+\frac{1}{4}[\cos \left({3x+720}^o\right)+3\cos ({240}^o+x)]$

$=\frac{1}{4}(\cos 3x+3\cos x)+[\cos 3x+3\cos ({120}^o+x)]+\frac{1}{4}[\cos 3x+3\cos ({240}^o+x)]$

$=\frac{3}{4}\cos 3x+\frac{3}{4}[\cos x+\cos ({120}^o+x)+\cos ({240}^o+x)]=\frac{3}{4}\cos 3x+\frac{3}{4}[\cos x+2\cos ({180}^o+x)+\cos {60}^o]$

$=\frac{3}{4}\cos 3x+\frac{3}{4}(\cos x-\cos x)=\frac{3}{4}\cos 3x$

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