$\sin \frac{1}{x} \neq \pm \frac{1}{x}$ for all real numbers $x$ excluding $0$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.