About the convexity of $\sin x$ for $\pi\leq x\leq 2\pi$ To prove the convexity of $\sin x$ over $[\pi,2\pi]$ through the second derivative is easy, but I would be interested in a (possibly) simple proof of convexity that avoids derivatives. Can you provide it?
 A: Since $\sin(x+\pi)=-\sin(x)$ and $\sin\left(\frac{\pi}{2}-x\right)=\sin\left(\frac{\pi}{2}+x\right)$, it is enough to prove that $\sin x$ is a concave function over $I=\left[0,\frac{\pi}{2}\right]$. The sine function is also continuous, hence it is enough to show midpoint-convexity over the same interval. Since:
$$\begin{eqnarray*} \sin(2a)+\sin(2b)-2\sin(a+b) &=&\text{Im}\left(e^{2ia}+e^{2ib}-2e^{i(a+b)}\right)\\&=&\text{Im}\left(e^{ia}-e^{ib}\right)^2\\&=&-4\sin^2\left(\frac{a-b}{2}\right)\,\sin(a+b)\end{eqnarray*} $$
for any $x,y\in I$ we have:
$$ \frac{\sin(x)+\sin(y)}{2}\leq\sin\left(\frac{x+y}{2}\right)\tag{1}$$
as wanted. 
We may notice that just like in the case of the exponential function, the definition through an ODE gives a way faster proof of convexity. Any solution of:
$$ f+f'' = 0$$
is concave when positive and convex when negative. 
On the other hand, a geometric proof of $(1)$ is quite straightforward:

In the above configuration, $OM$ and $AC$ are orthogonal, hence the area of $OAMC$ (i.e. twice the area of $AOM$) is greater than the area of $OABC$ (i.e. the area of $AOB$ plus the area of $BOC$).
