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?

Carrying your idea further, we have for $|x-y| < \delta(\epsilon) = \epsilon/2$
$$|\sin^2 x - \sin^2 y|= |\sin x + \sin y||\sin x - \sin y| \\ \leqslant 2 |\sin x - \sin y|\\ = 4\left|\sin\left(\frac{x-y}{2}\right)\right|\left|\cos\left(\frac{x+y}{2}\right)\right| \\ \leqslant 4\left|\sin\left(\frac{x-y}{2}\right)\right|\\ \leqslant 2|x-y| \\< \epsilon.$$
