# Showing $\frac{\sin a}{\sin b} <\frac{\pi}{2} \frac{a}{b}$ for $0<a<b<\frac{\pi}{2}$

I want to show that

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

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

• (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 $[0, \pi]$), $$0 < a < b < \frac\pi2 \implies \frac{\sin b}b > \frac{\sin(\pi/2)}{\pi/2} = \frac2\pi > \frac2\pi\frac{\sin a}a.$$ – Calum Gilhooley Feb 19 at 21:47
• In fact, the condition $a < b$ isn't needed, just $0 < a < \pi$ and $0 < b < \frac\pi2.$ – Calum Gilhooley Feb 20 at 6:10

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.

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

Also $$\sin x \leq x$$

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.