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I want to show that

$$\frac{\sin a}{\sin b} < \frac{\pi}{2} \frac{a}{b}$$ when $0<a<b<\frac{\pi}{2}$.

I tried to use MVT but I only got partial results... Does anyone know how to prove this ?

Thanks in advance.

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  • $\begingroup$ (This virtually duplicates user35508's answer, so I'm posting it as a comment.) By the inequality $\sin a < a$ ($a > 0$) and Aristarchus's inequality (user141614 gives a one-line proof using the strict concavity of the $\sin$ function on $[0, \pi]$), $$ 0 < a < b < \frac\pi2 \implies \frac{\sin b}b > \frac{\sin(\pi/2)}{\pi/2} = \frac2\pi > \frac2\pi\frac{\sin a}a. $$ $\endgroup$ – Calum Gilhooley Feb 19 at 21:47
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    $\begingroup$ In fact, the condition $a < b$ isn't needed, just $0 < a < \pi$ and $0 < b < \frac\pi2.$ $\endgroup$ – Calum Gilhooley Feb 20 at 6:10
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Since all the values are positive, this is equivalent to showing that

$$ \frac{\sin a}{a} <\frac{\pi}{2} \frac{\sin b}{b} $$

MacLaurin Series make the calculation fairly straight-forward, so let's ignore it. Instead, let's look at whether $\frac{\sin x}{x}$ is increasing on the interval which would finish the problem.

$$ \begin{align} \frac{d}{dx} \frac{\sin(x)}{x} &= \frac{x\cos x - \sin x}{x^2} \\ x \cos x & \stackrel{?}> \sin x \\ x & <\tan x \text{ for all } 0 < x < \frac{\pi}{2} \end{align} $$

Since $\frac{\sin x}{x}$ is decreasing between $0$ and $\frac{\pi}{2}$, we must hope that it doesn't decrease too much. The worst case scenario would be as $a \to 0$ and $b \to \frac{\pi}{2}$.

$$ \lim_{x \to 0} \frac{\sin x}{x} = \frac{\frac{d}{dx} \sin x}{\frac{d}{dx} x}\lim_{x \to 0} = \frac{\cos x}{1} = 1 \\ \frac{\pi}{2} \frac{\sin \frac{\pi}{2}}{\frac{\pi}{2}} = 1 $$

Which proves the inequality over the specified range.

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Use Jordan's inequality $\frac{2}{\pi} \leq \frac {\sin x }{x}$ Reciprocate this and use this for $b$

Also $\sin x \leq x$

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From the Maclaurin series expansion, you know that $\sin(x) < x$ for $x > 0$. Then it suffices to show that $$\frac{1}{\sin(x)} < \frac{\pi}{2x} \to \sin(x) > \frac{2x}{\pi}$$ for $0 < x < \frac{\pi}{2}$. Consider the function $f(x) = \frac{\sin(x)}{x}$. This approaches $1$ at $x = 0$ and is $\frac{2}{\pi}$ at $x = \frac{\pi}{2}$. So it is enough to show that $f'(x) < 0$ for $0 < x < \frac{\pi}{2}$. $f'(x)$ is equal to $\frac{x\cos(x)-\sin(x)}{x^2}$. Since $\sin(x) > x\cos(x)$ for $0 < x < \frac{\pi}{2}$, it follows that $f'(x) < 0$ for this interval. Thus $f(x)$ is decreasing on this interval and so must be between $1$ and $\frac{2}{\pi}$. Therefore, you have $$\sin(x) > \frac{2x}{\pi}$$ and the original inequality follows from there.

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