Chứng minh rằng $\left ( \frac{sinx}{x} \right )^3>cosx,\forall x \in\left ( 0;\frac{\pi}{2} \right )$
$\left ( \frac{sinx}{x} \right )^3>cosx,\forall x \in\left ( 0;\frac{\pi}{2} \right )$
#1
Đã gửi 16-06-2014 - 20:55
#2
Đã gửi 17-06-2014 - 01:29
Chứng minh rằng $\left ( \frac{sinx}{x} \right )^3>cosx,\forall x \in\left ( 0;\frac{\pi}{2} \right )$ (1)
(1) $\Leftrightarrow x^3<\frac{\sin^3x}{\cos x}=\tan x.\sin^2x=\tan x(1-\cos^2x)=\tan x-\frac{1}{2}\sin 2x$
Xét $f(x)=\tan x-\frac{1}{2}\sin 2x-x^3$ trên $\left(0;\frac{\pi}{2}\right)$
$f'(x)=\tan^2x+1-\cos 2x-3x^2$
$f''(x)=2\tan x(\tan^2x+1)+2\sin 2x-6x=2\tan^3x+2\tan x+2\sin 2x-6x$
$f'''=6\tan^2x(\tan^2x+1)+2(\tan^2x+1)+4\cos 2x-6=6\tan^4x+8\tan^2x+4\cos 2x-4$ $=6\tan^4x+8\tan^2x-8\sin^2x$
$=6\tan^4x+8\tan^2x\left(1-\cos^2x \right)\ge0,\ \forall x\in\left(0;\frac{\pi}{2}\right)$
$\Rightarrow \forall x\in\left (0;\frac{\pi}{2} \right)\ : \ f''(x)> f''(0)=0\Rightarrow f'(x)> f'(0)=0\Rightarrow f(x)> f(0)=0$
Suy ra đpcm.
#3
Đã gửi 05-09-2016 - 15:37
(\frac{sinx}{x})^3>cosx. \forall x \in (0; \frac{\pi}{2})
(\frac{sinx}{x})^3>cosx. \forall x \in (0; \frac{\pi}{2})
Xét g(x)=sinx-x
g'(x)=cosx-1>0 \forall x \in (0; \frac{\pi}{2})
Suy ra g(x)>g(0)=0
Đặt f(x)=sin^3x-x^3.cosx
Ta có:
f'(x)=3.sin^2x.cosx-(3x^2.cosx-x^3sinx)
f'(x)=3.cosx(sin^2x-x^2)+x^3sinx
f'(x)=3.cosx(sinx-x)(sinx+x)+x^3sinx>0 \forall x \in (0; \frac{\pi}{2})
Do cosx>0 ; sinx-x>0 ; sinx+x>0 ;x^3sinx>0 \forall x \in (0; \frac{\pi}{2})
Suy ra f(x)>f(0)
\Leftrightarrow sin^3x>x^3cosx
\Leftrightarrow \frac{sin^3x}{x^3}>cosx (Đpcm)
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