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I know there are some proofs showing that, for $0 < x < \pi/2$:

$\sin x < x < \tan x $

with the use of the unit circle, however, can I prove this algebraically in some way as well?

John_Average
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  • For some values of $x$ in $(0,2\pi)$, $\tan x$ is negative, which is not greater than $x$. To prove something algebraically, you also need to provide the definition with which you are working with. – player3236 Dec 02 '20 at 19:45
  • It depends on your definitions of trigonometric functions. If you know calculus then you can prove it using derivatives. However, sometimes you need these formulas before you can even compute the derivatives. It all depends on the definitions. – Mark Dec 02 '20 at 19:46
  • Presumably you mean $0 < x < \pi/2$. – Robert Israel Dec 02 '20 at 19:47
  • I'm pretty sure, these inequalities have already plenty of duplicates on this site, please use the search box. – zwim Dec 02 '20 at 19:48
  • Yes, I mean 0 < x < $\pi/2$. – John_Average Dec 02 '20 at 19:52
  • It's easy to show using derivatives, i.e. $(x-\sin x)'=1-\cos x \ge 0$ – Vasili Dec 02 '20 at 19:54
  • Since arclength can't (in general) be defined purely algebraically, you will need at least a minimal amount of calculus. You can get $\theta<\sin\theta$ by using the Pythagorean theorem and the fact a straight line is the shortest distance between two points. (Consider the points $(\cos\theta,\sin\theta)$, $(\cos\theta,0)$, $(1,0)$.) Getting $\theta<\tan\theta$ I'm not sure about - since arclength is a supremum of piecewise linear approximations, arclength seems amenable to lower bounds but not to upper bounds without more calculus. – anon Dec 02 '20 at 19:58
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    That depends on your definition of trigonometric functions. If you define it geometrically (unit circle), the obvious geometric proof would be most natural. If you use another method to introduce trigonometric functions (series, integrals, differential equations, functional equations,...), you may need other methods of proof. But you should clearly state your definitions, because otherwise, you risk circular reasoning. –  Dec 02 '20 at 20:05

2 Answers2

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One simple way could be to demonstrate that , over $[0, pi/2)$ $\sin$ is concave while $\tan$ is convex, and that $\sin(0)=0=\tan(0)$ and $\sin '(0) = 1 =\tan'(0)$.

G Cab
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Since $\sin' = \cos $ and $\cos' = -\sin$ and $0 < \cos < 1$ for $0 < x < \pi/2$, $\sin(x) =\int_0^x \cos(t) dt \lt x $.

Since $\tan'(x) =\sec^2(x) =\dfrac1{\cos^2(x)} $, $\tan(x) =\int_0^x\dfrac1{\cos^2(t)}dt \gt\int_0^x 1dt =x $.

marty cohen
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