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

Question

Prove that for $x\in\mathbb{R}$ $$ \cos x\geq 1-\frac{x^2}{2}. $$

My try:

Consider $g(x)=\cos(x)-1+\frac{x^2}{2}.$ If I differentiate $g(x)$ then we get $g'(0)>0$ so locally we get $g(x)>g(0)=0$ and then we can see that the function is increasing for any $x$ the function is increasing and hence we have $g(x)\geq 0$ for any $x \geq 0$. But I am getting that if $x<0$ then $g(x) \leq 0.$ So this inequality isn't true in general for all $x \in \Bbb R$.

But, if we use Taylor's theorem with Lagrange's remainder then also I am not sure what will be the point $\zeta\in [-x,0]$ where $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4}\cos(\zeta).$

Answer 1

You can use the following corollary of the Mean value theorem:

Let $f, g$ be differentiable functions on an interval $I$, $x_0\in I$, such that

1. $f(x_0)\ge g(x_0), $
2. $f'(x)\ge g'(x)$ $\:\forall x\ge x_0$.

Then $f(x)\ge g(x)$ $\:\forall x\in I, x\ge x_0$.

Now, as both sides of the inequality are even functions, it is enough to prove it for $x\ge 0$.

Let $f(x)=\cos x$, $\:g(x)=1-\dfrac{x^2}2$, $x_0=0$. We indeed have $f(0)=g(0)$, and $f'(x)=-\sin x$, $g'(x)=-x$, and it is well-known that $\sin x \le x$ on the interval $\bigl[0,\frac\pi 2\bigr]$, because $\sin x$ is concave on this interval. As $\frac\pi 2>1$, a fortiori, we have $\sin x \le x $ on $\mathbf R^+$. Therefore $$f'(x)=-\sin x\ge g'(x)=-x\qquad \forall x\ge 0$$

Answer 2

$$\cos x-1+\frac{x^2}2\ge0$$ and equality holds at $x=0$.

Then differentiating,

$$-\sin x+x\ge 0$$ and equality holds at $x=0$.

Finally,

$$-\cos x+1\ge 0.$$

So $-\sin x+x$ grows from $0$ and is non negative, and $\cos x-1+\dfrac{x^2}2$ grows from $0$ and is non-negative.

This technique works for Taylor developments to arbitrary orders.

Answer 3

we see that as $x\rightarrow -x$ we get the same inequality. Hence WLOG $x\ge 0$. AS $$\sin (x/2)\le \frac{x}{2} \tag1$$ $$1-2\sin^2 (x/2)\ge 1-\frac{x^2}{2}$$ $$\cos x \ge 1-\frac{x^2}{2}$$

Proof of Claim $(1)$ is very famous

Answer 4

The problem with using Taylor's theorem as you suggest is that the series only converges quickly near the origin, but we can restrict the range of $x$. If $|x|>2$ then $1-\frac{x^2}2<-1$ and the statement is true, so we may assume $|x|\leq2$. Then by Taylor's theorem,$$\begin{align} \cos x-1+\frac{x^2}2 &=\frac{x^4}{4!}-\frac{x^6}{6!}\cos\theta\\ &=\frac{x^4}{4!}\left(1-\frac{x^2}{30}\cos\theta\right)\\ &\geq\frac{x^4}{4!}\left(1-\frac4{30}\right)\geq0 \end{align}$$

Answer 5

Using the Lagrange remainder, $$ \tag1g(x)=\frac {x^4}4\cos \zeta\ge 0\qquad \text{for }|x|\le\frac\pi2$$ because that also make $|\zeta|<\frac\pi2$ and hence $\cos\zeta>0$. Also, we trivially have $$ g(x)\ge-2+\frac12x^2=\frac{x^2-4}2\ge 0\qquad\text{for }|x|\ge 2.$$ Hence we are only left with $\frac \pi2 <|x|<2$. But for these we already know from $(1)$ that $\cos \frac x2\ge 1-\frac18x^2>0$, so $$\cos x=2\cos^2\frac x2-1\ge 2\left(1-\frac18x^2\right)^2-1=1-\frac12x^2+\frac1{32}x^4\ge 1-\frac12x^2.$$

Answer 6

Note $$\frac{x^2}2-1 + \cos x =\int_0^x (t - \sin t)dt = \int_0^x dt\int_0^t (1-\cos s)ds \ge 0$$ Thus $$ \cos x\geq 1-\frac{x^2}{2} $$