Quadratic bound on $\sin^2(\theta)$

Question

Consider the the function $$ A \sin^2(\theta) $$ where $ A \in \mathbb{R}, ~ A > 0$ and $\theta$ is some angle.

I am working on a suboptimal control problem in which I am interested in controlling $\theta$. The particular method I am hoping to apply is valid if I can place bound on the nonlinearity that is quadratic in $\theta$. That is, I would need to show $$ A \sin^2(\theta) \leq B \theta^2 ~~ \forall ~ \theta $$ for some $B \in \mathbb{R}$.

My first thought is to let $A=B$ which leads to $$ \sin^2(\theta) \leq \theta^2 $$ I am not sure how to verify that this is true. Visually, if we plot both sides of the inequality on the same plot, it looks like the inequality is valid. However, looks like and proven to be are two very different things. We can take the square root of both sides and say that $$ | \sin(\theta) | \leq | \theta | $$ But I don't know how to take this any further and it is not clear to me how to show that the inequality holds, or not, for all $\theta$. Any nudges in the right direction are greatly appreciated.

Answer 1

Red: $y=|\sin x|$

Blue: $y=|x|$

Answer 2

$2\sin^2x=1-\cos(2x)$. The cosine series is alternating (for real $x$). And for all $|x|<1$ its terms are falling. Thus the expression is alternatingly bounded above and below by the partial sums of the series. In consequence $$ 2x^2-\frac43 x^4\le 1-\cos(2x)=2\sin ^2x\le 2x^2. $$ Obviously, for $|x|>1$ the inequality for the upper bound is trivial.