I came across a problem in section 3.4 on uniform continuity in Wade's 4th ed Introduction to Analysis. The exercise is just to apply the definition of uniform continuity to $f(x)=x\sin(2x)$ on $x\in(0,1)$.
The function is clearly uniformly continuous, which I don't have any problem showing. But I wanted to check on a step from the solution manual where there is an interesting global bound given: $|\sin(2x)-\sin(2a)|\leq 2|\sin(x-a)|$.
It is not stated as a global bound, but it seems to be one. I am able to derive it as a bound on $(0,1)$, but my sequence of steps seems too complicated for a basic undergrad advanced calc exercise.
My questions: Is this a well-known bound that most undergrads should have encountered? What is the quickest way to derive the bound, both globally and on $(0,1)$?
Here is my method for $(0,1)$. Assume $x,a\in(0,1)$ and $x>a$.
$$\begin{aligned} |x\sin(2x)-a\sin(2a)| &\leq|x\sin(2x)-x\sin(2a)| \qquad &\text{ (since $x>a$) }\\ &\leq|\sin(2x)-\sin(2a)| \qquad &\text{ (since $0<x<1$) }\\ &=|2\sin(x)\cos(x)-2\sin(a)\cos(a)| \qquad &\text{ (double angle identity) }\\ &\leq2|\sin(x)\cos(x)-\sin(a)\cos(x)| \qquad &\text{ ($x>a$ & $\cos$ decr.) }\\ &\leq2|\sin(x)\cos(a)-\sin(a)\cos(x)| \qquad &\text{ ($x>a$ & $\cos$ decr.) }\\ &=2|\sin(x-a)| \qquad &\text{ (angle difference identity) }\\ \end{aligned}$$
This argument uses the fact that $x,a\in(0,1)$ when I use the fact that $\cos(x)$ is decreasing on that interval and allows us to swap $\cos(x)$ and $\cos(a)$ in lines 4 and 5 in the above equation. Is there a quicker way to get from line 2 to line 6? It will need to be elementary, something that an undergrad in a very basic intro to analysis course might know. I suppose it could be the case that a sufficiently complicated argument is required and this just happens to be one of the harder problems.
Hint:
$\sin x - \sin y= $
$2 \cos\dfrac {x+y}{2} \sin \dfrac {x-y}{2}.$