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I need to prove that $\sin(x) > \frac{x}{2}$ if $0<x<\pi/2$

I've started working with the derivative, but if it's possible, I'd rather something simpler than that.

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Do you mean $\sin{x}<\frac{x}{2}$ if $0<x<\frac{pi}{2}$? I think you misspelled "sin" and you want a lower bound on $x$ or this isn't true...for example, $sin{\frac{-3\pi}{2}}=1$ is a counterexample. –  Logan Tatham Feb 12 at 19:11
I´m sorry, I meant sin, I keep writing it the spanish way... And I definitely meant $sin(x)>x/2$... Must miss the symbol while typing it. –  Luna Sage Feb 12 at 19:14
You mean when $0\lt x\lt \frac{\pi}{2}$. I find derivative easiest, one can use more basic but more complicated geometry. –  André Nicolas Feb 12 at 19:18
$\sin{x}$ is strictly concave on $x\in \left[0,\,\frac{\pi}{2}\right],$ therefore, $\sin{x}>\frac{2}{\pi}x>\frac{1}{2}x.$ –  M. Strochyk Feb 12 at 19:28
Ok, then I´ll keep working with the derivate, but I have problems when I consider cos(x) after $\pi/6$, beacuse it reaches 1/2, and that is the derivate of x/2... any ideas? –  Luna Sage Feb 12 at 19:29

4 Answers 4

up vote 8 down vote accepted

Let's rewrite the desired inequality as

$${\theta\over2}\lt\sin\theta\quad\text{for }0\lt\theta\le{\pi\over2}$$

To prove this, draw the portion of the unit circle in the first quadrant, draw a typical angle $\theta$ from the origin $O$ to a point $P=(\cos\theta,\sin\theta)$, and let $Q=(1,0)$. The wedge of the circle for this angle has area $\theta/2$. It consists of two pieces: the triangle $\triangle OPQ$, of area ${1\over2}\sin\theta$, and a lens-shaped piece outside the triangle.

Here's the key point to convince yourself of: If you reflect the lens-shaped region across the line $PQ$, it stays inside the triangle $\triangle OPQ$, and therefore has smaller area. The upshot is that the area of the wedge is less than twice the area of the triangle, which is exactly what we want.

Added later: Here's a second proof based on the same picture.

In addition to the points $P=(\cos\theta,\sin\theta)$ and $Q=(1,0)$, let $R=(1,y)$ be the point of intersection of the two tangents to the circle at $P$ and $Q$.

Now the length of the circular arc from $Q$ to $P$ is, by definition, $\theta$ (when the angle is measured in radians). But this length is less than the sum of the lengths $QR$ and $PR$. By symmetry, those two lengths are the same. Thus, since the length of $PR$ is obviously just $y$, we have $\theta\lt2y$. But $R$ clearly lies closer to the $x$-axis than $P$, hence $y\lt\sin\theta$.

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Thanks so much! –  Luna Sage Feb 12 at 20:01
@LunaSage, you're welcome. I added a second proof just for fun. –  Barry Cipra Feb 12 at 20:15
Both the proofs are almost falling in category of "proof without words" if a diagram is given. Its bit tricky compared to usual proof of $\sin \theta < \theta$. Really very clever. +1 –  Paramanand Singh Feb 13 at 4:50

Since $\sin x > x-\frac{x^3}{6}$ for $x>0$, we can show that $x-\frac{x^3}{6}>x/2$ with your condition. That is, $$x>\frac{x^3}{3}\\1>\frac{x^2}{3}\\ \text{At the max of x, show } 1>\pi^2/12$$

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Since $\sin(x)$ is concave on $[0,\pi]$, we have $$ \begin{align} \sin(x) &\ge\sin(0)+(x-0)\frac{\sin(\pi/2)-\sin(0)}{\pi/2-0}\\ &=\frac{2x}{\pi}\\ &\ge\frac x2 \end{align} $$

$\hspace{8mm}$enter image description here

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Which theorem are you using? –  Luna Sage Feb 12 at 19:54
Or that´s part of the definition? –  Luna Sage Feb 12 at 19:57
@LunaSage: I am using the fact that $\sin(x)$ is concave and that for a concave function $$f(x)\ge f(a)+(x-a)\frac{f(b)-f(a)}{b-a}$$ on $[a,b]$. –  robjohn Feb 12 at 20:15
Thanks so much! I can only accept one answer, I think, but this is really useful. +1 –  Luna Sage Feb 13 at 17:35
@LunaSage: Glad it was helpful. Yes, you can upvote or downvote each of the answers, but you can only accept one. –  robjohn Feb 13 at 18:00

Draw the line trough $(\cos x,\sin x)$ and tangent to the unit circle. This line slopes down (at an angle $x-\tfrac{\pi}{2}$) and intersects the line $x=1$ in the point $(1, \tan \tfrac{x}{2})$. So $$ \sin x>\tan\frac{x}{2}>\frac{x}{2}.$$

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