Theorem 1   The sine function is concave on the interval $[0, \pi]$ .

Proof. Suppose that $x$ and $y$ lie in the interval $[0, \pi/2]$ . Then $\sin x$ , $\sin y$ , $\cos x$ , and $\cos y$ are all non-negative. Subtracting the identities $$ \sin^2 x + \cos^2 x = 1 $$ and $$ \sin^2 y + \cos^2 y = 1 $$ from each other, we conclude that $$ \sin^2 x - \sin^2 y = \cos^2 y - \cos^2 x . $$ This implies that $\sin^2 x - \sin^2 y \ge 0$ if and only if $\cos^2 y - \cos^2 x \ge 0$ , which is equivalent to stating that $\sin^2 x \ge \sin^2 y$ if and only if $\cos^2 x \le \cos^2 y$ . Taking square roots, we conclude that $\sin x \le \sin y$ if and only if $\cos x \ge \cos y$ .

Hence, we have $$ (\sin x - \sin y) (\cos x - \cos y) \le 0 . $$ Multiply out both sides and move terms to conclude $$ \sin x \cos x + \sin y \cos y \le \sin x \cos y + \sin y \cos x . $$ Applying the angle addition and double-angle identities for the sine function, this becomes $$ {1 \over 2} \left( \sin (2x) + \sin (2y) \right) \le \sin (x + y) . $$ This is equivalent to stating that, for all $u,v \in [0, \pi]$ , $$ {1 \over 2} \left( \sin u + \sin v \right) \le \sin \left( {u + v \over 2} \right) , $$ which implies that $\sin$ is concave in the interval $[0, \pi]$ .