concavity of sine function

Theorem 1.

The sine function is concave on the interval $\mathrm{[}\mathrm{0}\mathrm{,}\pi \mathrm{]}$.

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

$$\frac{1}{2}\left(\sin (2x)+\sin (2y)\right)\le \sin (x+y).$$

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

$$\frac{1}{2}\left(\sin u+\sin v\right)\le \sin \left(\frac{u+v}{2}\right),$$

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

Title	concavity of sine function
Canonical name	ConcavityOfSineFunction
Date of creation	2013-03-22 17:00:26
Last modified on	2013-03-22 17:00:26
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	8
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 26A09
Classification	msc 15-00