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Use the mean value theorem to prove that:

$$\cos(x)>1-\frac{x^2}{2}$$ for all $$x>0$$

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2 Answers 2

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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.

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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.

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