Why is $\sin(x^{2})$ similar to $\sin(x) \cdot x$? Why is $\sin(x^2)$ similar of  $\ x \sin(x)$? 
I graphed it using desmos and when I look at it, the behavior as x approaches zero seems to be to oscillate less. 
Yet as x approaches infinity and negative infinity $\sin(x^2)$ oscillates between y=1 and y=-1 while $\ x *sin(x)$ oscillates between y=x and y=-x.
I was wondering why these functions are so similar yet so different. I'm in 10th grade and I"m currently learning precalculus so if answers could be targeted to a precalculus level that would be great.
 A: When $x$ gets close to zero, $\sin x \approx x-\frac{x^3}{6}$. So $$\sin(x^2)\approx x^2-\frac{x^6}{6}\\x\sin(x)\approx x^2-\frac{x^4}{6}$$
Now, when $x$ is small, $x^4$ and $x^6$ are "very small." So the functions are dominated by $x^2$ near $x=0$. Indeed, if you graphed $y=x^2$ alongside, you'd see that both of your functions are close to  but smaller than $y=x^2$.
Add in $y=\sin^2(x)$, and it will be similar, too.
When you get to calculus, this will be explored by studying "power series" for functions.
We also see from this approximation that since $\frac{x^6}{6}<\frac{x^4}{6}$ when $x$ near enough to zero (say $|x|<1$) we see that $\sin(x^2)>x\sin(x)$.
A: In calculus, you learn that many functions can be written as "infinite polynomials." For example, there's this:
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dotsb$$
Yes, those are factorials. A consequence of this is that, since $\sin\pi=0$, we have:
$$0=\pi-\frac{\pi^3}{3!}+\frac{\pi^5}{5!}-\frac{\pi^7}{7!}+\dotsb$$
This is not obvious.
Note that it begins with $x$, rather than $2x$ or $\frac x2$ or whatever. One consequence of this is that, when $x\approx0$, we have $\sin x\approx x$ (since all of the other terms are much smaller than $x$ when $x$ is small). This is actually much easier to prove than the rest of the series; the usual proof involves some geometry and the unit circle. However, for what follows, the exact coefficient of $x$ doesn't really matter.
In any case, multiplying by $x$, we get:
$$x\sin x=x^2-\frac{x^4}{3!}+\frac{x^6}{5!}-\frac{x^8}{7!}+\dotsb$$
Or, replacing the $x$s with $x^2$s:
$$\sin x^2=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+\dotsb$$
We can see that these both begin with $x^2$. As $x$ gets smaller, these other terms are much less significant than the $x^2$ term. This means that, when $x\approx0$, we have $x^2\approx x\sin x\approx\sin x^2$.
A: Try finding the $ \lim_{x \rightarrow 0} \frac{\sin(x^2)}{x \sin(x)}$, you will see it is finite, thus they behave simiraly near $0$, which is important for approximations near $0$. 
Although $\sin(x^2)$ is bounded, and $x \sin(x)$ is not, they have local minima and maxima at the same points.
A: This is not an answer but it is too long for a comment.
You received the explanation of what you observed.
I give you another one which is also amazing : function $y=\sin^x(x)$ in the range $(2\pi,3\pi)$ almost looks like a gaussian. Doing things similar as in the answers, consider Taylor expansion around $x=\frac{5\pi} 2$ $$x\log(\sin(x))=-\frac{5\pi}{4}   \left(x-\frac{5 \pi }{2}\right)^2+O\left(\left(x-\frac{5 \pi
   }{2}\right)^3\right)$$ $$\sin^x(x)\simeq e^{-\frac{5\pi}{4}   \left(x-\frac{5 \pi }{2}\right)^2}$$ Pu the two functions on the same plot and enjoy.
