Show that $\frac{x}{\sin x}$ is strictly increasing on $(0,\frac{\pi}{2})$ I have tried by calculating the derivative, $$f'(x)=\frac{\sin x-x\cos x}{\sin^2 x}$$ However I don't know how to show that  $ \frac{\sin x}{x} > \cos x $. If this approach won't work, how else sould I tackle this problem?
 A: your Statement is equivalent to $$\frac{\sin(x)}{\cos(x)}=\tan(x)>x$$ which is true. 
Hint: consider $$f(x)=\tan(x)-1$$
A: Inequality $\tan{x}>x$ holds on $(0,\frac{\pi}{2})$.
Proof of this fact you can check here.
A: $\begin{array}\\
\dfrac{\sin x}{x}
&=\dfrac1{x}\int_0^x \cos(t)dt\\
&\gt\dfrac1{x}(x\cos(x))
\qquad\text{since }\cos(t)\text{ is decreasing on }[0, \pi/2]\\
&=\cos(x)\\
\end{array}
$
A: Overkill: we have
$$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right),\qquad \cos(x) = \prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2\left(n-\frac{1}{2}\right)^2}\right) \tag{1} $$
and for every $x\in[0,\pi/2]$
$$ \left(1-\frac{x^2}{\pi^2 n^2}\right)\geq \left(1-\frac{x^2}{\pi^2\left(n-\frac{1}{2}\right)^2}\right)\tag{2} $$
holds for every $n\geq 1$.
A: We may as well prove that ${\rm sinc}(x):={\sin x\over x}$ is decreasing  on $\left[0,{\pi\over2}\right]$. Now for all $x\geq0$ we have
$${\rm sinc}(x)=\int_0^1\cos (t\,x)\>dt\ ,$$
and this is clearly decreasing in $x$ as long as $0\leq x\leq{\pi\over2}$.
A: *

*$x$ is strictly increasing since it's slope is $1$

*$\sin(x)$ is strictly increasing over the interval since it's slope is $\cos(x)$, which is positive in this interval.

*For all $x$ in the interval the interval, $1\ge\cos(x)$.


Therefore, the $\frac{x}{\sin(x)}$ is strictly increasing​ on the interval.
A: Suppose $0\lt y\lt x\lt\pi$. Then
$$
\begin{align}
\frac{x\sin(y)-y\sin(x)}{x-y}
&=\frac{(x-y)\sin(y)-y\,(\sin(x)-\sin(y))}{x-y}\\
&=y\left(\frac{\sin(y)-\sin(0)}{y-0}-\frac{\sin(x)-\sin(y)}{x-y}\right)\\[6pt]
&\ge0\tag{1}
\end{align}
$$
because $\sin(x)$ is concave on $[0,\pi]$.
Multiplying $(1)$ by $\frac{x-y}{\sin(x)\sin(y)}$, we get
$$
\frac{x}{\sin(x)}-\frac{y}{\sin(y)}\ge0\tag{2}
$$
That is, $\frac{x}{\sin(x)}$ is increasing on $(0,\pi)$.

In the answer above, we can avoid the use of calculus, if we accept that $\sin(x)$ is concave.
However, we can also note that on $[0,\pi]$,
$$
\sin''(x)=-\sin(x)\le0\tag{3}
$$
We can also use the fact that $\cos(x)$ is decreasing on $[0,\pi]$ and the Mean Value Theorem to get that
$$
\underbrace{\frac{\sin(y)-\sin(0)}{y-0}}_{\cos(u)\text{ for some } u\in(0,y)}-\underbrace{\frac{\sin(x)-\sin(y)}{x-y}}_{\cos(v)\text{ for some } v\in(y,x)}=\cos(u)-\cos(v)\ge0\tag{4}
$$
since $u\lt v$.
