Is $\sin^2x$ uniformly continuous on$x\in [0,\infty]$ I have the question that is $sin^2x$ uniformly continuous on $x \in [0,\infty]$ ?
My approach:
Let $\left|x-y\right|<\delta$ we have:-
$$\left|sin^2x-sin^2y\right|=\left|(\sin x+\sin y)(sin x-sin y)\right|$$
$$=4\left|\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)\right|\lt4\left|\sin \left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right|\tag{1}$$ 
$$\lt \left|(x-y)(x+y)\right|<\left|(x+y)\right|\delta,$$which is dependent on $x$ so $\sin^2\!x$
is not uniformly continuous.
Is this solution correct or not? I have some doubt about validity of inequality $(1)$ also, if it is correct then why?

I'm not sure if you are "allowed" to use this method, but for a place to start:
Note that the derivative $2\sin x\cos x$ is bounded between $[-2,2]$
Now we want to show that for any $\epsilon$, we can pick $\delta$ such that $|x-y|<\delta$ implies $|\sin^2x-\sin^2y|<\epsilon$.
And since the derivative is bounded, we have that for all $x,y$ $$\left|\frac{\sin^2x-\sin^2y}{x-y}\right|\le 2$$
And you can probably take it from here.
