$\sin \frac{1}{x} \neq \pm \frac{1}{x}$ for all real numbers $x$ excluding $0$ Question: Prove that for all real numbers $x$ excluding $0$, $\sin(\dfrac{1}{x}) \neq \pm\dfrac{1}{x}$. 
Here's what I've done so far:
Let $y = \dfrac{1}{x}$. Then $y\in (-\infty,0)\bigcup(0,\infty)$. We know that for $0 < y < \dfrac{\pi}{2}, -1<0<\cos y<\dfrac{\sin y}{y}<1 \Rightarrow -y<\sin y < y$. Since $\dfrac{\sin(-y)}{-y} = \dfrac{\sin y}{y}$ and $\cos(-y)=\cos y$, then for $-\dfrac{\pi}{2} < y<0, -1<0<\cos y<\dfrac{\sin y}{y} < 1 \Rightarrow -y>\sin y > y$. Since $\max(\sin y) = 1$ and $\min(\sin y) = -1 \forall y\in \mathbb{R}, y \in (-\infty,0)\bigcup(0,\infty)$, when $y>1$, $\sin y \neq y\wedge \sin y \neq -y$ (notice that if we let $y < -1,$ we will obtain an equivalent expression). So, considering all cases, $\sin y \neq y, -y \forall y\in \mathbb{R}, y \in (-\infty,0)\bigcup(0,\infty)$. Since $y= \dfrac{1}{x},$ then $sin(\dfrac{1}{x})\neq \dfrac{1}{x} \forall x\in \mathbb{R}, x\neq 0$
Edit: I don't know if it's the work above that makes this question confusing, but I'll delete my "work" if need be. It should be very obvious as to what the question is.
 A: Proving that $\sin(\frac{1}{x})\neq \pm\frac{1}{x}\forall x\neq 0$ is same as saying that the only possible solutions of $y=\pm\sin y$ are at $0$, which is true, because the functions $y-\sin y$ and $y+\sin y$ are non-decreasing on $\mathbb R$ (that you can check by differentiation) and therefore intersect $x$-axis only once.
A: The  function $$y=\sin t$$ and functions  $$y= \pm t$$ intersect only at $t=0$
Let $t= 1/x$ and the result is $$\sin( 1/x) = \pm  1/x$$ do not have any  real solution.
A: Here is a geometric proof that $\sin\theta\neq\theta$ for $0<\theta<\frac\pi2 $ (measured in radians, of course).

Consider an angle of measure $\theta$ centered on the unit circle.  $CD=\sin\theta$ and the length of arc $CB$ is $\theta$.  Since the unique shortest path from a point to a line is the perpendicular line segment from the point to the line, we conclude that $\sin\theta<\theta$.
The proofs considering the functions are stronger because they consider the entire domain of $x\neq0$, but perhaps this gives you a visual intuition of why it would be true near zero.
