Verify $\cos{x}<\left(\frac{\sin{x}}{x}\right)^3$ for $0<x<\pi/2$.

Question

Verify by Mean Value Theory or otherwise that $\cos{x}<\left(\frac{\sin{x}}{x}\right)^3$ for $0<x<\pi/2$.

I am unable to solve the problem. Please give me a solution of the problem.

Answer 1

We need to prove that $f(x)>0$, where $f(x)=\frac{\sin{x}}{\sqrt[3]{\cos{x}}}-x$.

Indeed, by AM-GM $$f'(x)=\frac{\cos{x}\sqrt[3]{\cos{x}}+\frac{\sin^2x}{3\sqrt[3]{\cos^2x}}}{\sqrt[3]{\cos^2x}}-1=\frac{3\cos^2x+\sin^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}=$$ $$=\frac{1+\cos^2x+\cos^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}\geq\frac{3\sqrt[3]{\cos^4x}-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}=0.$$ Thus, $f(x)>f(0)=0$ and we are done!

Answer 2

For $x > 0$ the well-known representation as alternating series give the estimates $$ \sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} > x - \frac{x^3}{6} \, ,\\ \cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} < 1 - \frac{x^2}{2} + \frac{x^4}{24} \, . $$

For $0 < x < \pi/2$ in particular $x^2 < 6$ holds and therefore $$ \left( \frac{\sin x}{x} \right )^3 - \cos x > \left( 1 - \frac{x^2}{6} \right )^3 - \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} \right ) \\ = \frac{x^4}{24}- \frac{x^6}{216} = \frac{x^4}{24} \left( 1 - \frac{x^2}{9} \right) > 0 \, . $$

Answer 3

By brutal force method, let consider

$$f(x)=\left(\frac{\sin{x}}{x}\right)^3-\cos{x} \implies f'(x)=\frac{\sin x (x^4 + 3 x \cos x \sin x - 3 \sin^2x)}{x^4}=\frac{\sin x}{x^4}h(x)$$

and

$$h(x)=x^4 + 3 x \cos x \sin x - 3 \sin^2x \implies h'(x)=4x^3+3x\cos (2x)-\frac32\sin (2x)$$

$$\implies h''(x)=6x(2x-\sin (2x))> 0$$

with $h(0)=0$ therefore $f'(x)>0$ and then $f(x) >0$ since at $x \to 0$ $f(x)=\frac{x^4}{15}+O(x^6)$.