Question

${\cos }^{3}x+{\cos }^3({120}^0-x)+{\cos }^3({120}^0+x)$ lies in the interval

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

Answer 1

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

On expanding and using the compound angle formula, we get
$cosx+\cos (x+{120}^0)+\cos ({120}^0-x)=0$ - (i)

We have,
$a^3+b^3+c^3=3abc$ if $a+b+c=0$ - (ii)

Let $S={\cos }^{3}x+{\cos }^3({120}^0-x)+{\cos }^3({120}^0+x)$
$\Rightarrow S=3\cos x\times \cos ({120}^0-x)\times \cos ({120}^0+x)$ (From (i), (ii) )
$\Rightarrow S=3\cos x\times (\frac{-1}{2}\cos x+\frac{\sqrt{3}}{2}sinx)\times (\frac{-1}{2}\cos x-\frac{\sqrt{3}}{2}sinx)$
$\Rightarrow S=\frac{3}{4}(4{\cos }^{3}x-3\cos x)$

$\Rightarrow S=\frac{3}{4}\cos \left(3x\right)$

$\cos 3x$ can acquire values from [$-1,1$]
Hence, the range of $S$ is [ $\frac{-3}{4},\frac{3}{4}$]