I want to show that
$$\frac{\sin a}{\sin b} < \frac{a}{b} < \frac {\tan a} {\tan b}$$ when $0<b<a<\frac{\pi}{2}$.
I notice that $\sin x < x < \tan x$ on this interval. So maybe I could study the behavior of the function $f(x,y)=x/y$? Or is there a simpler way to approach this?
$f(x)={{sin(x)}\over x}$, $f'(x)={{xcos(x)-sin(x)}\over x^2}$
consider $g(x)=xcos(x)-sin(x), g'(x)=cos(x)-xsin(x)-cos(x)=-xsin(x)< 0$ in $(0,\pi/2)$ implies that $g$ strictly decreases, since $g(0)=0$, we deduce that $g(x)< 0$ and $f$ strictly decreases. This implies that for $b<a$, ${{sin(b)}\over b}> {{sin(a)}\over a}$ and ${{sin(a)}\over{sin(b)}}<{a\over b}$.
Alternative approach: $f(x)=\frac{x}{\sin x}$ is an analytic function in a neighbourhood of the origin. It is an even function and the radius of convergence of its Taylor series and the origin equals $\frac{\pi}{2}$. Additionally, all the coefficients of such Taylor series are non-negative and depend on the values of the $\zeta$ function at $2,4,6,\ldots$, since
$$ \frac{\sin z}{z} = \prod_{n\geq 1}\left(1-\frac{z^2}{n^2\pi^2}\right) $$
and we may apply $\frac{d}{dz}\log(\cdot)$ or $\frac{d^2}{dz^2}\log(\cdot)$ to both sides of such identity (Weierstrass product).
The same applies to $g(x)=2-\frac{x}{\tan x}$. It follows that both $f(x)$ and $g(x)$ are non-negative, increasing and convex over $\left(0,\frac{\pi}{2}\right)$. That is enough for proving the given inequality. Actually, we have much more: both $f(x)$ and $g(x)$ are log-convex over the same interval.
We should mention that the function $\frac{x}{\sin x}$ is strictly increasing on $[0, \pi)$, while the function $\frac{\tan x}{x}$ is strictly increasing on $[0, \frac{\pi}{2})$. This can be proved by calculating derivatives. It it interesting that for both functions, their Taylor series at $0$ have positive coefficients and converge on the intervals $(-\pi, \pi)$, respectively $(\frac{\pi}{2}, \frac{\pi}{2})$.
