Bài 104720

Question

Chứng minh rằng:

$\forall x \in (0,\frac{\pi}{2})$
a.

$\sin x<x$
b.

$\sin x >x -\frac{x^{3}}{6}$
c.

$( \frac{\sin x}{x} ) ^{3} >\cos x$

score 0 · Answer 1

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)

1 Đáp án