Prove $1-\frac{1}{2} x^{2} \leq \cos x,\: \forall x \in \mathbb{R}$

Question

Prove that

$$1-\frac{1}{2} x^{2} \leq \cos x,\: \forall x \in \mathbb{R}$$

I am trying to prove this using taylor's theorem:

We have: $$f(x)=f(0)+f'(0)x+\frac{x^2}{2!}f''(0)+...$$ Now approximating the function as a polynomial of degree $3$ we get: $$\cos x=1-\frac{x^2}{2!}+\frac{\sin c}{6}x^3$$ where $c \in (0, x)$

Case $1.$ if $0 < x \le \pi$, then we have $\sin c \geq 0$ and $x^3 >0$, so we get $$\cos x \geq 1-\frac{x^2}{2}$$

Case $2.$ If $\pi < x \leq 2\pi$, i am confused, because $\sin c<0$ Any help?

A sketch often helps. If $|x|>2$ then $1-\frac12x^2 <-1\le \cos(x)$ — Henry, Oct 10 '21 at 13:48
Because of symmetry of both sides respect to the line $x=0$ it is enough to prove for $x\ge 0$. The function $f(x)=\cos(x)-1+\dfrac {x^2}{2}$ is increasing and $f(0)=0$ so $f$ is positive in the first quadrant. — Piquito, Oct 10 '21 at 14:09
https://math.stackexchange.com/questions/1513020/prove-the-inequalities-1-fracx22-le-cosx-le1-fracx22-fracx4 — Albus Dumbledore, Oct 10 '21 at 15:07
@AlbusDumbledore: You are right, I had failed to find that before. I think we can close the question as a duplicate. The other Q&A has several good answers. — Martin R, Oct 10 '21 at 17:05
Does this answer your question? Prove the inequalities $1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$ — Martin R, Oct 10 '21 at 17:05

Martin R · Accepted Answer · 2021-10-10T14:24:49.933

2

I am trying to prove this using Taylor's theorem ....

That is possible, just use the remainder for the term with exponent $2$, i.e. $$ f(x)=f(0)+f'(0)x+f''(c) \frac{x^2}{2!} $$ for some $c$ between $0$ and $x$. This gives $$ \cos(x) = 1 + 0 \cdot x + \underbrace{(-\cos(c))}_{\ge -1} \cdot \frac{x^2}{2!}\ge 1 - \frac{x^2}{2} $$ for all $x \in \Bbb R$.

edited Oct 10 '21 at 14:24

answered Oct 10 '21 at 14:15

Martin R

113,040

Brilliant proof. But how about my approach? can you help with case $2$? – Umesh shankar Oct 10 '21 at 14:19
@Umeshshankar: Your Taylor formula works only if $\sin(c) \ge 0$ is known. In the second case you can argue that $1-\frac12x^2 <-1\le \cos(x)$, as mentioned in a comment. – Martin R Oct 10 '21 at 14:22

score 1 · Answer 2 · answered Oct 10 '21 at 13:52

1

Consider a function $$f(x) =\cos x -1 + 2^{-1} x^2 $$ for $x\geq 0.$ Since $f'(x) =\sin x +x \geq 0$ thus the function is monotone increasing and therefore $$f(x) \geq f(0)=0$$ for $x\geq 0,$

but $f(-x) =f(x) $ and hence $$f(x) \geq f(0)=0$$ for all $x\in \mathbb{R}.$

answered Oct 10 '21 at 13:52

So is it not possible to prove using $\sin c$ term? – Umesh shankar Oct 10 '21 at 14:00

score 1 · Answer 3 · answered Oct 10 '21 at 14:13

If $x=0$, then the inequality is correct.

Let $x≠0$, then we have

$$\begin{align}&1-\frac{1}{2} x^{2}≤ \cos x\\ \iff &\frac{1-\cos x}{2}≤\frac 14 x^2\\ \iff &\sin^2\left(\frac x2\right)≤\frac {x^2}{4} \\\iff &\left|\sin\left(\frac x2\right)\right|≤\left|\frac x2\right|\\ \iff &|\sin t|≤|t|\\ \iff&\left|\frac{\sin t}{t}\right|≤1,\thinspace t≠0.\end{align}$$

This is the well-known fact that,

$$-1≤\cos x ≤ \frac{\sin x}{x} ≤ 1 $$

This completes the proof.

Note: The downvote is of no value as long as it is for no reason. — lone student, Oct 13 '21 at 19:21

score 1 · Answer 4 · edited Oct 11 '21 at 20:31

My answer's approach resembles that of MotylaNogaTomkaMazura but since his $f'[x]$ is wrong and I can't comment on his answer, I hope my answer will give you a better idea. Let $$f[x] = \cos[x] - 1 + \frac{1}{2}x^2$$ for $x \ge 0$ $$f'[x] = x - \sin[x]$$ We can prove $f'[x] \ge 0$ by taking its derivative: $$f''[x] = 1 - \cos[x] \ge 0$$ since $\cos[x] \le 1$. Since $f'[0] = 0$ and $f''[x] \ge 0, f'[x] \ge 0.$ Then again, $f[0] = 1 - 1 + 0 = 0$ and $f'[x] \ge 0$; therefore $f[x] \ge 0$ and thus $$\cos[x] \ge 1 - 1/2 x^2$$ Since $f[x]$ is an even function, $$\cos[x] \ge 1 - 1/2 x^2$$ for all $x$.

Thomas Andrews · Answer 5 · 2021-10-10T14:57:01.303

This proof uses that $|\sin y|\leq |y|.$

Consider the right triangle $A=(\cos x,0),$ $B=(\cos x,\sin x)$ and $C=(1,0).$

Then $$|BC|^2=(1-\cos x)^2+\sin^2 x=2-2\cos x.$$

But we can also show that $|BC|=2|\sin(x/2)|,$ by considering it as the base of an isosceles triangle, $OBC,$ with angle $x$ and equal sides $1.$

We also know that $|\sin y|\leq |y|.$

So $$2-2\cos x=4\sin^2\frac x2<x^2.$$

Rearranging gives your inequality.

This is almost entirely a geometric proof, except for the sine inequality we use.

We can show $|\sin y|\leq |y|$ geometrically, too, if we accept that the shortest path between two points is a line.

The line segment between $(\cos y,\sin y)$ and $(\cos y,-\sin y)$ is of length $2|\sin y|,$ while the path along the circle between the two points is length $2|y|.$ The circle path has to be at least as long as the shortest path, and we get our sine inequality.

Prove $1-\frac{1}{2} x^{2} \leq \cos x,\: \forall x \in \mathbb{R}$

5 Answers5