# Prove that: $\sin(x) \cos(x) \geq x-x^3$

Prove that the following inequality holds for $x\ge0$ :

$$\sin(x) \cos(x) \geq x-x^3$$

This is an inequality often met during my high school classes I also used for this problem yesterday. I'm interested in a non-calculus proof if this is possible.

Proof involving calculus:

Let's consider

$$f(x) = \sin(x) \cos(x)-x+x^3$$ then $$f'(x) = 3 x^2-2\sin^2(x)\tag1$$ $$x\ge \sin(x)\tag2$$ From $(1)$ and $(2)$ we immediately notice that $f'(x)\ge0$ and taking into account that $f(0)=0$ we may conclude that the inequality holds. Thanks.

-
is not good to multiply both side by 2?you get $sin(2*x)>2*(x-x^3)$ –  dato datuashvili Aug 23 '12 at 8:34
@dato: it could be. I'll check that. –  Chris's sis Aug 23 '12 at 8:36
wolframalpha.com/input/?i=sin%282*x%29%3E2*%28x-x%5E3%29 –  dato datuashvili Aug 23 '12 at 8:38
Maybe it helps that from $\sin x<x$ and the fact that for $0<x\le1$ we have $\sin x>0$ we get $\sin^2 x<x^2)$ (for $0<x\le1$, we can just square the inequality, and beyond we just notice that $\sin^2 x\le 1 < x$). Therefore we get $\cos^2 x = 1-\sin^2 x > 1-x^2$ and thus $x\,\cos^2 x > x-x^3$. However I have no idea how to prove without calculus that $x\,\cos x < \sin x$ for $x<\pi/2$ (you can see that by plotting both functions). –  celtschk Aug 23 '12 at 9:32

This is a completely revamped proof. We only use the two inequalities $$x \geq \sin(x) \text{ and } \tan(x) \geq x$$ which have elementary proofs here.

Case 1: $0 \leq x < \pi/2$

We know that $x \geq \sin(x)$. This gives us $x^2 \geq \sin^2(x)$, for with $x \leq \pi$ both terms are positive and simply squaring is justified, and with $x \geq \pi$, $x^2 \geq \pi^2 \geq 1 \geq \sin^2(x)$.

We can do some manipulations on this inequality to find an inequality involving cos: $$x^2 \geq \sin^2(x)\\ x^2 \geq 1 - \cos^2(x)\\ \cos^2(x) \geq 1- x^2$$ We now have $$\sin(x) \cos(x) = \frac{\sin(x)}{\cos(x)} \cos^2(x) = \tan(x) \cos^2(x) \geq x(1- x^2) = x-x^3$$ as required.

Case 2: $\pi/2 \leq x$

The previous argument only works for small $x$, where tan is well behaved. For $\pi/2 \leq x$ we see that $x-x^3 \leq \pi/2 - (\pi/2)^2 \leq -1$ (as the function $g(x) = x-x^3$ is decreasing for $x \geq 1$), and as $-1 \leq \sin(x) \cos(x)$ (as both sin and cos are large than -1), the result follows.

-
what did happen with your previous post? –  Chris's sis Aug 23 '12 at 10:47
@Chris'sister - I'm making a nicer proof. –  James Fennell Aug 23 '12 at 10:49
that's good! Thank you! –  Chris's sis Aug 23 '12 at 10:51
@ James Fennell: your way seems really nice and simple at the same time! –  Chris's sis Aug 23 '12 at 10:52
Nice answer (+1)! Both of these inequalities, $\sin(x)\le x\le\tan(x)$, are shown in this geometric proof, too. –  robjohn Aug 23 '12 at 23:20

A Geometric Proof

The posed inequality is equivalent to $\sin(x)\ge x-x^3/4.$

Consider the wedge of the unit circle below:

$\hspace{32mm}$

The area of the whole wedge (red and green regions) is $\frac12x$, and the area of the green triangle is $\frac12\sin(x)$. Thus, we get that $\sin(x)\le x$. Furthermore, the area of the red region is $\frac12(x-\sin(x))$.

Noting that the red region is contained in the rectangle with base $2\sin(x/2)$ and height $1-\cos(x/2)$, we get that \begin{align} \tfrac12(x-\sin(x)) &\le2\sin(x/2)(1-\cos(x/2))\\ &=4\sin(x/2)\sin^2(x/4)\\ &\le x^3/8 \end{align} which yields $$x-x^3/4\le\sin(x)$$ as desired.

-
amazingly beautiful! This proof will be clearly part of my favorite proofs-collection! (+1) and a big Thank you! –  Chris's sis Aug 23 '12 at 22:26
how did you think of this approach? :) I love to see limits solved by geometrical approaches. It's simply fascinating. –  Chris's sis Aug 23 '12 at 22:35
@Chris'sister: This pretty much follows the same thinking that went into this similar answer. –  robjohn Aug 23 '12 at 22:37
That's nice to see, too. –  Chris's sis Aug 23 '12 at 22:41
What software did you use for making this drawing? I'd be interested in some free software to make such drawings... –  Chris's sis Sep 8 '12 at 19:18

As others have already noticed, multiplying both sides by $2$ makes it sufficient to prove $\sin 2x \ge 2x - 2x^3$ or $\sin x \ge x - \frac{x^3}{4}$.

One method of proving this is the following:

It is well known that $\sin x \le x$, therefore

$$\int_0^x (t - \sin t) \, dt \ge 0,$$

giving us that $\cos x \ge 1 - \frac{x^2}{2}$.

Applying the same thing again, we have

$$\int_0^x \cos t \, dt \ge \int_0^x \left(1 - \frac{t^2}{2}\right) \, dt$$

giving us that $\sin x \ge x - \frac{x^3}{6}$, so in fact we've proven a stronger inequality!

Note: It is easy to see that this method yields easily that the Taylor's series stopped at positive terms (negative terms) overestimates (respectively, underestimates) $\sin x, \cos x$.

-
Integration is calculus, too, isn't it? –  celtschk Aug 23 '12 at 9:37
Well, a positive function having a positive area below it seemed much more intuitive than anything else I could think of.. –  Rijul Saini Aug 23 '12 at 9:39

Multiply both sides by $2$ and then by drawing curves of the two functions, namely $\sin 2x$ and $x^3-x$, find the region where the curve of $\sin x$ is above the other in the graph.

-
(+1) for your efforts. –  Chris's sis Aug 23 '12 at 8:57
I fixed your answer by using $\LaTeX$ but you reverted the changes. I don't quite get the reason. Feel free to edit it yourself. –  Gigili Aug 23 '12 at 9:00
@Gigili : It looked as if there was a serious typo in your edit. –  Michael Hardy Aug 23 '12 at 9:54
Then BruinJ came along and wrote sin $2x$ instead of $\sin 2x$. –  Michael Hardy Aug 23 '12 at 9:56
@MichaelHardy: Well, seems I forgot a dollar sign and it was a bit messed up. –  Gigili Aug 23 '12 at 10:07

I can't seem to come up with a non-calculus method of solving, but taking @dato's trig trick, you can multiply both sides by $2$ yielding:

$$2\sin x\cos x\geq 2(x-x^3)$$

$$\sin2x\geq 2x-2x^3$$

And using a Taylor expansion:

$$2x-\frac{8x^3}{3!}+\mathcal{O}(x^5)\geq 2x-2x^3$$ $$\frac{12x^3}{6}-\frac{8x^3}{6}+\mathcal{O}(x^5) = \frac{4x^3}{6}+\mathcal{O}(x^5)\geq 0$$ ... when $x\geq 0$.

-
(+1) for your efforts. –  Chris's sis Aug 23 '12 at 8:56
A Taylor expansion solution seems like a calculus-based solution to me... –  Potato Aug 23 '12 at 9:12
Correct, as indicated in the first line. I was just showing a straightforward, "alternative" calculus approach. –  user12345121212 Aug 23 '12 at 15:19
Once again, and despite the upvotes, one should mention that limited expansions CANNOT yield such inequalities, for any fixed $x\gt0$. –  Did Aug 23 '12 at 16:32