Use the mean value theorem to prove that:
$\cos(x)>1-\frac{x^2}{2}$ for all $x>0$
Use the mean value theorem to prove that:
$\cos(x)>1-\frac{x^2}{2}$ for all $x>0$
I thought there may be better way than my answer... But I'll stick on my answer.
apply mean value theorem on $ f(x)= \frac{x^2}{2}+ \cos x $
since $f(0)=1$, we have some $ x_0 \in (0,x)$ such that
$ \frac{\frac{x^2}{2}+ \cos x -1}{x} = x_0 - \sin x_0 >0$
since $x>0$, we get the results.
okay...so this is my contribution
let $h(x)=cos(x)+((x^2)/2)$
the derivative of $h(x)=x-sin(x)$
for all x>0, the derivative of h is greater than 0.
i.e. $x-sin(x)>0$
let a,b be subsets of x>0, then there exist an number c such that the derivative of c is
$(f(b)-f(a))/b-a.$
let b=x and a=0 from the interval x>0.
hence, we have $(f(x)-f(0))/x-0=(cos(x)+((x^2)/2)-1-0)/x-0=$the derivative of c but the derivative of c is greater than 0. therefore, (cos(x)+((x^2)/2)-1)/x>0 (cos(x)+((x^2)/2)-1)>0 this implies that cos(x)=1-(x^2)/2).PROVED.