Prove that

**dhiab** · August 16th 2009, 12:27 PM

Prove that : $\forall x > 0:1 - \frac{{x^2 }}{2} \le \cos x \le 1<br /> <br />$

**red_dog** · August 16th 2009, 01:19 PM

We have to prove the first inequality. The second is obviously.

Let $f(x)=1-\frac{x^2}{2}-\cos x, \ x>0$ .

$f'(x)=-x+\sin x$

$f''(x)=-1+\cos x\leq 0, \forall x\in\mathbb{R}\Rightarrow$ f' is decreasing.

Then, $\forall x>0\Rightarrow f'(x)\leq f'(0)=0\Rightarrow$ f is decreasing.

Then, $\forall x>0\Rightarrow f(x)<\leq f(0)=0\Rightarrow 1-\frac{x^2}{2}\leq \cos x$