Using mean value theorem to show that $\cos (x)>1-x^2/2$

$ f(x)=x^2/2+\cos(x)$

Note that $f(0)=0^2/2+1=1$ From your equation: $$f(x)-f(0)=(x)f'(c)=x(c-\sin c)$$

Let $g(x)=x-\sin x$ Again you can show that $g'(x)=1-\cos x$ which is always greater than $0$ due to bounded nature of $\cos x$.As $g(0)=0$ and it is an increasing function $\{g'(x)>0\;\forall x>0$}, thus $g(x)>0 \;\forall x>0$.

So $f(x)-f(0)>0\;\forall x>0$ as $x>0$ and $c-\sin c >0\;\forall c>0${as $0<c<x$}.

So $f(x)>f(0)=1$

You started off well.

Notice that, by MVT:

$$f'(c) = \frac{f(x) - f(0)}{x - 0}$$ S0

$$xf'(c) = f(x) - f(0)$$

Notice that x is positive, and since $$f'(x) = x - sin(x)$$

Also, note that $x > \sin(x)$, so $f'(x) > 0$

Therefore,

We can conclude that

$$f(x) > f(0)$$

And

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