For $0 < \theta < {\pi}/{2} $, $$0 < \cos \theta < \frac{\sin \theta}{\theta} < \frac{1}{\cos \theta}$$ I understand that proving this requires an advanced knowledge of calculus, so I was looking for the geometric intuition behind this. The only part that is really surprising to me is $\cos \theta < \frac{\sin \theta}{\theta}$, which reduces to $\tan \theta > \theta$ (since $\tan \theta = \frac{\sin\theta}{\cos\theta}$).

I know that the trigonometric functions can be represented like this: https://i.stack.imgur.com/3Cfhl.jpg (really beautiful geomtry here) so that the question really is to prove that the line segment $\tan \theta$ is longer than the arc $\theta$.

Notice that if you drop a perpendicular to the $x$-axis from the point where the radius touches the circle, the area under $\tan \theta$ is always greater than that under $\theta$ (for $0 < \theta < \pi/2$).

I am outlining one way to prove that $\tan \theta > \theta$: Prove that if the area under some curve $f(x)$ equals the area under another curve $g(x)$ for some respective $x_1$ and $x_2$, the lengths of the two curves will be equal, too (which sounds reasonable yet surprising at the same time). Since the area under $\tan\theta$ is clearly greater than that under $\theta$, $\tan \theta > \theta$.

Could someone explain why (if) my hypothesis about the areas and lengths of curves is true? I'm looking for an intuitive rather than necessarily rigorous proof (or contradiction). Other surprising ways to prove the inequality are welcome too.

  • $\begingroup$ The statement "if $$\int_{x_1}^{x_2} f dx = \int_{x_1}^{x_2} g dx$$ then the arc-length of $f$ from $x_1$ to $x_2$ equals the arc-length of $g$ from $x_1$ to $x_2$" is false. Consider for example, $x_1 = 0, ~ x_2 = 2 \pi$, and the two curves $f(x) = 0$ and $g(x) = \sin x$. $\endgroup$ – paulinho May 30 '20 at 15:42
  • $\begingroup$ @paulinho You misunderstood me. $x_1$ is not the lower bound on both integrals. Some constant is (say $0$). $x_1$ is the upper bound for the integral of $f$ and $x_2$ the upper bound for that of $g$. $\endgroup$ – Ray Bradbury May 30 '20 at 15:53
  • $\begingroup$ Do you accept that the area of a unit circle sector of angle $\theta$ is $\frac12\theta$? $\endgroup$ – user May 30 '20 at 15:55
  • $\begingroup$ @IshanJMukherjee I see. But even this is not true. Take the lower limit of both integrals to be zero, and now take $x_1 = \infty$ and $x_2 = 1$. Let $f = e^{-x}$ and $g = 1$. The arc-length of $g$ is clearly finite from $0$ to $1$, but the arc-length of $f$ over the given interval is infinite. Nevertheless, both areas under the curve are equal to $1$. $\endgroup$ – paulinho May 30 '20 at 15:57
  • $\begingroup$ @paulinho I'm curious: Is there a contradiction if you allow only curves that have finite arc length? If so, do you have any other way of answering my original question? $\endgroup$ – Ray Bradbury May 30 '20 at 16:00

Obviously (see figure) $$[\triangle AOC]<[\sphericalangle AOC ]<[\triangle AOB]\tag1$$ where $[\dots]$ means the area.

Assuming the radius of the circle $OA=1$ this amounts to $$ \frac12\sin\theta<\frac12\theta<\frac12\tan\theta,\tag2 $$ where $\theta= \measuredangle AOC$.

enter image description here

  • $\begingroup$ I accepted at the outset that the area under $\tan \theta$ is greater than that under $\theta$. Does this also necessarily mean that $\tan \theta$ is greater than $\theta$? If yes, can it be geometrically shown? If not, is there some other way to show that $\tan \theta > \theta$ $\endgroup$ – Ray Bradbury May 30 '20 at 16:39
  • $\begingroup$ The figure shows geometrically that the areea of the circular sector is less than that of triangle $OAB$. And the areas are $\frac12\theta$ and $\frac12\tan\theta$, respectively. From $x/2<y/2$ follows $x<y$. $\endgroup$ – user May 30 '20 at 16:43
  • $\begingroup$ You are right, I was being careless. I have accepted your answer. $\endgroup$ – Ray Bradbury May 30 '20 at 16:47

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