Question

Find the maximum value of ${\cos }^{3}x+{\cos }^3(120-x)+co{s}^3(120+x)$.

Answer 1

Given: ${\cos }^{3}x+{\cos }^3(120-x)+co{s}^3(120+x)=$ $\frac{\cos 3x+3\cos x}{4}+\frac{\cos (360-3x)+3\cos (120-x)}{4}+\frac{\cos (360+3x)+3cos(120+x}{4}$
[Since, $\cos 3x=\frac{\cos 3x+3\cos x}{4}$]
= $\frac{1}{4}(\cos 3x+3\cos x+\cos 3x+3\cos (120-x)+\cos 3x+3\cos (120+x)$
= $\frac{3}{4}(\cos 3x+\cos x+\cos (120-x)+\cos (120+x)$
= $\frac{3}{4}(\cos 3x+\cos x+2\cos 120\cos x)$
=$\frac{3}{4}(\cos 3x+\cos x+2\times (-\frac{1}{2}\cos x\left)\right)$
= $\frac{3}{4}(2\cos 2x\cos x-\cos x)$
=$\frac{3}{4}(\cos 3x+\cos x-\cos x)$
$\therefore$${\cos }^{3}x+{\cos }^3(120-x)+co{s}^3(120+x)=$ $\frac{3}{4}\cos 3x$ and
Maximum value of $\cos 3x=1$.
So, maximum value of ${\cos }^{3}x+{\cos }^3(120-x)+co{s}^3(120+x)=$ $\frac{3}{4}\times 1=\frac{3}{4}$