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For all $0<\alpha<\frac{\pi}{6}$

Prove that $\frac{\sin(\alpha)}{\alpha}>\frac{3}{\pi}$

MY ATTEMPT :

Since $$\frac{1}{\cos\alpha}>\frac{\sin\alpha}{\alpha}>\cos\alpha$$

Therefore, $$\frac{\sin\alpha}{\alpha}>\cos\alpha$$

Since $\cos\alpha$ decreases in $0<\alpha<\frac{\pi}{6}$

we can deduce that $$\frac{\sin\alpha}{\alpha}>\frac{\sqrt3}{2}$$

But $\frac{\sqrt3}{2}<\frac{3}{\pi}$

So the minimum is not $\frac{3}{\pi}$

Any methods to prove

Angelo Mark
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  • You proved that $\sqrt 3/2$ is a lower bound, that does not mean that $\sqrt 3/2$ is the minimum, but that does neither disprove it. – Christophe Boilley May 12 '25 at 13:25
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    "Since $a>b>c,$ therefore $b>c$" is a weird deduction. Not wrong, but it seems redundant, like you seem to mean something else? – Thomas Andrews May 12 '25 at 13:33
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    See https://math.stackexchange.com/q/213382/42969 for a similar estimate (“Jordan's inequality”). All those proofs can be adapted to the interval $[0, \pi/6]$. – Martin R May 12 '25 at 14:34

3 Answers3

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Proof by derivative

$$f(x)=\frac{\sin x}{x}$$ $$f'(x)=\frac{\color{red}{x\cos x}-\sin x}{x^2}<\frac{\sin x-\sin x}{x^2}=0$$ Therefore $f$ is decreasing on $(0,\frac{\pi}{6}]$, so it's minimum occur at the boundary $$f\left(\frac{\pi}{6}\right)=\frac{3}{\pi}$$

Proof by integral representation

Because $\cos x$ is decreasing on $(0,\frac{\pi}{6}]$, we have $$\frac{\sin x}{x}=\int_{0}^{1}\cos(tx)\,\mathrm dt\ge\int_{0}^{1}\cos\left(t\cdot\frac{\pi}{6}\right)\,\mathrm dt=\frac{3}{\pi}$$

Thinh Dinh
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8

Let $P\equiv (\alpha,\sin \alpha)$ be point on the graph of the curve $y=f(x)=\sin x$, where $0<\alpha<\frac{\pi}{6}$

Also let $O\equiv (0,0)$ and $A\equiv (\frac{\pi}{6},\frac{1}{2})$

Since $f''(x)<0$ for $x\in (\frac{\pi}{6},\frac{1}{2})$ we have

Slope of OP must be greater than slope of OA i.e$$m_{OP}>m_{OA}$$$$\frac{\sin\alpha-0}{\alpha-0}>\frac{\sin\frac{\pi}{6}-0}{\frac{\pi}{6}-0}$$$$\frac{\sin\alpha}{\alpha}>\frac{\frac{1}{2}}{\frac{\pi}{6}}$$$$\frac{\sin\alpha}{\alpha}>\frac{3}{\pi}$$

Maverick
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3

$$f(t)=\sin t -\frac{3t}{\pi},\ f''(t)=-\sin t <0, $$for all $t\in (0, \frac{\pi}{6})$. Therefore $f$ is concave on this interval. Since $f(0)=f(\frac{\pi}{6})=0$ we have $f(t)>0$ for all $t\in (0, \frac{\pi}{6})$.

Letac Gérard
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