Is $\cos(x) \geq 1 - \frac{x^2}{2}$ for all $x$ in $R$? I encountered this question:
Is $\cos(x) \geq 1 - \frac{x^2}{2}$   for all $x$ in  $R$  ?
I draw to myself the function graphs but i can't find the right way to answer this type of questions.
Cosx function graph
1-x^2/2 function graph
 A: Another method is using the well known inequality $\sin { x\le x } $ and integrating both sides $$\int _{ 0 }^{ x }{ \sin { tdt\le \int _{ 0 }^{ x }{ tdt }  }  } $$ 
$${ \cos { t } | }_{ 0 }^{ x }\le { \frac { { t }^{ 2 } }{ 2 } | }_{ 0 }^{ x }$$
$$-\cos { x } +1\le \frac { { x }^{ 2 } }{ 2 } \\$$

$$ \cos { x } \ge 1-\frac { { x }^{ 2 } }{ 2 } $$

A: Consider the definition of $\cos x$ as the sum of a power series:
$$\cos x=\sum_{k\ge0}(-1)^k\frac{x^{2k}}{(2k)!}.$$
As $\cos x$ is an even function, and the series has terms of even degree only, we may suppose $x\ge 0$.
Furthermore, we may suppose  $x\le 2$ because $1-\frac{x^2}2<-1$ for $x>2$. For each  $x\in[0,2]$, it's an alternating series with  terms decreasing in magnitude. Partial sums are approximations of $\cos x$ with an error
$$\cos x-\biggl(\sum_{k=0}^n(-1)^k\frac{x^{2k}}{(2k)!}\biggr)= (-1)^{n+1}\frac{x^{2(n+1)}}{(2(n+1))!}+\dotsm$$
which has the sign of the first omitted term (and its absolute value is not more than the absolute value of this term).
Hence $\;1-\dfrac{x^2}{2}<\cos x$.
A: $1-\frac{x^2}{2}=-1\iff x=\pm2$ so we verify the inequality just on $[-2,2]$.
The function $$f(x)=\cos x-(1-\frac{x^2}{2})=\cos x +\frac{x^2}{2}-1$$ is even ($f(-x)=f(x)$) so it is enough to verify $f(x)$ is positive in $[0,2]$.
Take the derivative $$f'(x)=-\sin x+x$$ we know that $f'(x)\gt 0$ because the line $y=x$ is tangent at $x=0$ to the curve $g(x)= \sin x$ which is concave downward at positive neighborhood of $0$. Hence $f(x)$ is increasing on $[0,2]$
Thus $f(x)\ge 0$ (actually  $f(x)\gt 0$ for $x\ne 0$)
A: Consider the function
$$
f(x)=\cos x - 1 + \frac{x^2}{2}
$$
We have $f(0)=0$ and
$$
f'(x)=x-\sin x
$$
with $f'(0)=0$. Also
$$
f''(x)=1-\cos x
$$
which shows that $f'(x)$ is a strictly increasing function, because its derivative is positive except on a set of isolated points (that has no limit point). Therefore $f'(x)>0$ for $x>0$ and $f'(x)<0$ for $x<0$. Hence $0$ is an absolute minimum for $f$. Since $f(0)=0$, we have $f(x)>0$ for every $x\ne0$. This means that, for every $x$,
$$
\cos x\ge 1-\frac{x^2}{2}
$$
equality holding only for $x=0$.
