How to study the monotonicity of $f(x) = \sin 2x$, without using differentiation and continuity.

Question

How to study the monotonicity of $f(x) = \sin 2x$, $x \in [-\pi, \pi]$, without using differentiation and continuity.

Answer 1

It is : $ -1 \leq \sin x \leq 1$, $\forall x\in\mathbb R$. $(1)$

You have : $\sin(-\pi)=\sin(\pi)=\sin(0) = 0$ and also $\sin(-\pi/2) = -1,\sin(\pi/2) = 1$. $(2)$

Now, you know that $f(x) = \sin x$ is a continuous function $\forall x \in \mathbb R$. Taking into account this and $(1),(2)$, you can conclude that $f$ is decreasing in the intervals $[-\pi,-\pi/2],[\pi/2,\pi]$ and increasing in the intervals $[-\pi/2,\pi/2].$

Answer 2

The sinus addition formula gives

$$\sin(2x)-\sin(2y)=2\cos(x+y)\sin(x-y)$$

But first let's notice that $f(x)=\sin(2x)$ has period $\pi$.

So we can study on $[0,\pi]$ and to make it even more simple, let's shift the interval by $-\frac{\pi}4$ and study for $x\in[\frac{-\pi}4,\frac{3\pi}4]$.

case 1 : $-\frac{\pi}4\le x\le y\le \frac{\pi}4$

$-\frac{\pi}2\le x+y\le\frac{\pi}2\implies\cos(x+y)\ge 0\quad$ and $\quad-\pi\le x-y\le 0\implies\sin(x-y)\le 0$

So $\quad\sin(2x)\le\sin(2y)\quad$ and $\quad f$ is $\nearrow$ on this interval.

case 2 : $\frac{\pi}4\le x\le y\le \frac{3\pi}4$

$\frac{\pi}2\le x+y\le\frac{3\pi}2\implies\cos(x+y)\le 0\quad$ and $\quad -\pi\le x-y\le 0\implies\sin(x-y)\le 0$

So $\quad\sin(2x)\ge\sin(2y)\quad$ and $\quad f$ is $\searrow$ on this interval.

conclusion:

Completing by $\pi$ periodicity you get $\begin{cases} f \nearrow \text{ on }[-\pi,-\frac{3\pi}4]\cup[-\frac{\pi}4,\frac{\pi}4]\cup[\frac{3\pi}4,\pi]\\ f \searrow \text{ on }[-\frac{3\pi}4,-\frac{\pi}4]\cup[\frac{\pi}4,\frac{3\pi}4]\\ \end{cases}$