a)Xét: $f(x)=x-\sin x$
$f'(x)=1-\cos x>0,\forall x \in (0,\frac{\pi}{2})$
$\Rightarrow f(x)>f(0)=0,\forall x \in (0,\frac{\pi}{2})$
Vậy:$x>\sin x ,\forall x \in (0,\frac{\pi}{2})$
b)Xét: $g(x)=\sin x -x +\frac{x^{3}}{6}$
$g'(x)=\cos x -1+\frac{x^{2}}{2}$
$=-2\sin^{2} \frac{x}{2}+\frac{x^{2}}{2}=\frac{1}{2}[(\frac{x}{2})^{2}-(\sin \frac{x}{2})^{2}]>0$
(do câu a)
$\Rightarrow g(x)>g(0),\forall x \in (0,\frac{\pi}{2})$
Mà $g(0)=0$.Vậy: $\sin x >x -\frac{x^{3}}{6},\forall x \in (0,\frac{\pi}{2})$
c)Theo (b): $( \frac{\sin x}{x} ) ^{3}>( 1 -\frac{x^{2}}{6}) ^{3}$
Mà:$0<x<\frac{\pi}{2} \Rightarrow x^{2}<9$
$\Rightarrow \frac{x^{4}}{24}>\frac{x^{6}}{216}$
$\Rightarrow 1-\frac{x^{2}}{2}+\frac{x^{4}}{12}-\frac{x^4}{24}<1-\frac{x^{2}}{2}+\frac{x^{4}}{12}-\frac{x^{6}}{216}=(1-\frac{x^{2}}{6})^{3}$
$\Rightarrow 1-\frac{x^{2}}{2}+\frac{x^{4}}{24}<( \frac{\sin x}{x} ) ^{3} (1)$
Hơn nữa:
Xét $h(x)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\cos x$
$\Rightarrow h'(x)=-x+\frac{x^{3}}{6}+\sin x=\sin x-(x-\frac{x^{3}}{6})$
$\Rightarrow h''(x)=cosx-1+\frac{x^2}{2}$
$h'''(x)=-sinx+x>0, \forall x\in(0;\frac{\pi}{2})$ (theo câu a)
$\Rightarrow h''(x)>h''(0)=0,\forall x \in (0,\frac{\pi}{2})\Rightarrow h'(x)>h'(0)=0$
$\Rightarrow h(x)>h(0)=0, \forall x\in(0;\frac{\pi}{2})$
$\Rightarrow \cos x<1-\frac{x^{2}}{2}+\frac{x^{4}}{24}, \forall x \in (0,\frac{\pi}{2}) (2)$
Từ (1) và (2) $\Rightarrow ( \frac{\sin x}{x} ) ^{3} >\cos x $
$\Rightarrow$ (ĐPCM)