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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Since $\sin(x)$ is concave on $[0,\pi]$, we have $$ \begin{align} \sin(x) &\ge\sin(0)+(x-0)\frac{\sin(\pi/2)-\sin(0)}{\pi/2-0}\\ &=\frac{2x}{\pi}\\ &\ge\frac x2 \end{align} $$
$\hspace{8mm}$
robjohn
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4@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 '14 at 20:15
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Thanks so much! I can only accept one answer, I think, but this is really useful. +1 – Lessa121 Feb 13 '14 at 17:35
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@LunaSage: Glad it was helpful. Yes, you can upvote or downvote each of the answers, but you can only accept one. – robjohn Feb 13 '14 at 18:00
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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 $y\lt\sin\theta$.
Barry Cipra
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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 '14 at 04:50
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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, $$x>\frac{x^3}{3}\\1>\frac{x^2}{3}\\ \text{At the max of x, show } 1>\pi^2/12$$
Flowers
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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 $$ \sin x>\tan\frac{x}{2}>\frac{x}{2}.$$
WimC
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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 '14 at 19:11