Intersections of trigonometric functions and $x$

Question

I was fiddling with my calculator and disovered something odd: $\sin x$ only intersects $x$ (as it seems) at $x=0$. Why is that? Furthermore, what is the significance of the intersection of $\cos x$ and $x$ (about 0.7390851332)?

Answer 1

The function $f(x) = \sin x$ has $f(0) = 0$, and its gradient is always <= 1 (as $f'(x) = \cos x$). In order to meet the simple function $g(x) = x$ anywhere else, its gradient would have to be > 1 somewhere, or else =1 everywhere (by the Intermediate Value Theorem).

I don't believe the point $\cos x = x$ has any major significance, as the geometrical meaning of cos is that x is an angle (whose numerical value depends on what units you use) whereas $\cos x$ is a ratio of two lengths (which doesn't depend on units).

Answer 2

Let $f(x)=x-\sin x$. Then $\forall x, f'(x)=1-\cos x\ge0$ as $-1\le\cos x\le 1$.

Therefore the curve either curves upward ($f'(x)>0$) or levels off/remains constant. That implies that for every $x$, $f(x)\le f(x+\delta x)$. So to proof that $x=0$ is the only solution, one way is to show that between $-\delta x$ and $\delta x$, $f(x)$ is increasing but not remaining constant.

For $f'(x)=0,x=2k\pi, k\in{\Bbb{Z}}$. So by taking $\delta x=2\pi$, we have for $-2\pi<x<2\pi$, $x=0$ is the only point that $f'(x)=0$; while $\forall x, 0<|x|<2\pi$, $f'(x)>0$, which is increasing. So for any $x\neq0, f(x)\neq f(0)=0$.

We can now conclude $x=0$ is the only solution.

Wolfram Alpha: https://www.wolframalpha.com/input/?i=x-sinx

Answer 3

The $x$-coordinates of the points of intersection of the curves $y = \sin x$, $y = \cos x$ with the line $y = x$, that is, $x$ such that $x = \sin x$ or $x = \cos x$, do indeed have a geometrical interpretation of some interest, which I shall momentarily discuss; but first, let us dispense with the problem of the unicity of $0$ as the solution to $x = \sin x$:

Set

$\sigma(x) = x - \sin x; \tag{1}$

we observe that $x \ge \pi/2$ implies

$\sigma(x) = x - \sin x \ge \dfrac{\pi}{2} - 1 > 0, \tag{2}$

using the fact that $\sin x \le 1$ everywhere. Now if $x \in (0, \pi/2]$, we may write

$\sigma(x) = \sigma(x) - \sigma(0) = \int_0^x \sigma'(t)dt = \int_0^x (1 - \cos t) dt > 0, \tag{3}$

since

$1 - \cos t > 0 \tag{4}$

for $t \in (0, \pi/2]$. We conlude from (2) and (3) that $\sigma(x)$ has precisely one zero for $x \in [0, \infty)$, and that is $x = 0$.

For $x \le 0$ the corresponding conclusion may be drawn by virtue of the fact that $\sigma(x)$ is an odd function, $\sigma(-x) = -\sigma(x)$, viz.

$\sigma(-x) = -x - \sin (-x) = -x + \sin x = -\sigma(x). \tag{5}$

We thus see that $0$ is the unique zero of $\sigma(x)$, i.e., $0$ is the only solution to $x = \sin x$.

A similar argument shows that the number $\alpha = .739085 \ldots$ is the only solution to $x = \cos x$; taking

$\tau(x) = x - \cos x, \tag{6}$

we see that

$\tau(-\dfrac{\pi}{2}) = -\dfrac{\pi}{2} \tag{7}$

and

$\tau(\dfrac{\pi}{2}) = \dfrac{\pi}{2}; \tag{8}$

it now follows from the intermediate value theorem that $\tau(x)$ has a zero somewhere in $(- \pi / 2, \pi / 2)$; furthermore,

$\tau'(x) = 1 + \sin x > 0 \tag{9}$

for $x \in (-\pi / 2, \pi / 2)$; thus the graph of $\tau (x)$ intersects the $x$-axis precisely one time in this interval, and this we have seen at $\alpha$; $\tau(x)$ has no other zeroes, since $\vert x \vert > \vert \cos x \vert$ for $\vert x \vert \ge \pi /2$. So $x = \cos x$ has the unique solution $\alpha$.

These results submit to geometrical interpretation: $x$ being measured in radians, an angle of size $x$ subtends an arc of length $x$ on a circle of unit radius; thus $\alpha$ may be thought of as that unique angle, measured in a counter-clockwise sense from the point $(1,0)$ on the unit circle, at which the arc-length along the circle from $(1, 0)$ (again taken in the counter-clockwise direction) is equal to the $x$-coordinate of the point on the circle corresponding to $\alpha$, that is, of $(\cos \alpha, \sin \alpha)$. And if arc-length in the clockwise direction is also admitted, then $(\cos \alpha, -\sin \alpha)$ also shares this property. Likewise, $(1, 0)$ is the only point whose $y$-coordinate gives its arc-distance; but the central angle now must vanish, as we have seen; the arc-distance 'twixt $(1, 0)$ and itself is indeed $0$.