# concavity of sine function

###### 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 $\sin^{2}x-\sin^{2}y\geq 0$ if and only if $\cos^{2}y-\cos^{2}x\geq 0$, which is equivalent  to stating that $\sin^{2}x\geq\sin^{2}y$ if and only if $\cos^{2}x\leq\cos^{2}y$. Taking square roots, we conclude that $\sin x\leq\sin y$ if and only if $\cos x\geq\cos y$.

Hence, we have

 $(\sin x-\sin y)(\cos x-\cos y)\leq 0.$

Multiply out both sides and move terms to conclude

 $\sin x\cos x+\sin y\cos y\leq\sin x\cos y+\sin y\cos x.$

Applying the angle addition and double-angle identities for the sine function, this becomes

 ${1\over 2}\left(\sin(2x)+\sin(2y)\right)\leq\sin(x+y).$

This is equivalent to stating that, for all $u,v\in[0,\pi]$,

 ${1\over 2}\left(\sin u+\sin v\right)\leq\sin\left({u+v\over 2}\right),$

which implies that $\sin$ is concave in the interval $[0,\pi]$. ∎

Title concavity of sine function ConcavityOfSineFunction 2013-03-22 17:00:26 2013-03-22 17:00:26 rspuzio (6075) rspuzio (6075) 8 rspuzio (6075) Theorem msc 26A09 msc 15-00