Prove $\frac{\pi}{\sin^2 \frac{\pi}{x}} \gt \frac{x}{\tan \frac{\pi}{x}}$

Question

Not much explanation needed beyond the heading. I am trying to prove this above statement: $$\frac{\pi}{\sin^2 \frac{\pi}{x}} \gt \frac{x}{\tan \frac{\pi}{x}}$$ is true for all $$x>2$$ Any proofs or suggestions on how to get going would be appreciated.

The inequality is not defined for all real numbers (that's not an answer, just an observation) — Naj Kamp, Sep 01 '20 at 16:40
@NajKamp, I actually need it true for X>2, I have edited this and really should of put it in when I posted, I'm new at this. I think it will be defined for all x>2. — Scuffed Newton, Sep 01 '20 at 16:50
What have you tried? What failed? What do you think will work? — RSpeciel, Sep 01 '20 at 17:55

mjw · Accepted Answer · 2020-09-01T16:59:05.527

2

We want $$\frac{\pi}{\sin^2 \frac{\pi}{x}} \gt \frac{x}{\tan \frac{\pi}{x}}$$

When the denominators are non-zero, we can rewrite this as:

$$\frac{\pi}{x} > \sin \frac{\pi}{x} \cos \frac{\pi}{x}$$

Well, for all $p\ne 0$, $$|p|>|\sin p| \ge |\sin p \cos p|.$$

The expression is true for all $x \ne 0 \text{ (and where defined).}$

edited Sep 01 '20 at 16:59

answered Sep 01 '20 at 16:44

mjw

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Okay, yes, you are correct. I need to be careful because diving by $-4$ flips the sign of the inequality. – mjw Sep 01 '20 at 16:54
but then $\sin x > x$ using expansions.. so it would be true for all values of $|x|\geq2$, values of $0\leq |x|<2$ since $\sin x$ and $\tan x$ would become zero at some values// – Safdar Faisal Sep 01 '20 at 16:58
YEs! The expression $\pi - x \sin \pi/x \cos \pi/x $ is even, so you are right! – mjw Sep 01 '20 at 17:01

score 1 · Answer 2 · answered Sep 01 '20 at 16:56

$$ \frac \pi {\sin^2 \frac\pi x} > \frac x {\tan \frac{\pi}{x}} $$ Let $u = \pi/x.$ then $x>2$ means $0<u<\pi/2.$ Then the inequality becomes $$ \frac u {\sin^2 u} > \frac 1 {\tan u} \quad\text{for } 0<u<\frac \pi 2 $$ and then $$ u \tan u > \sin^2 u \quad \text{for } 0<u<\frac\pi2. $$ Since $\sin u < u < \tan u$ for $0<u<\pi/2,$ the inequality follows.

score 0 · Answer 3 · answered Sep 01 '20 at 16:59

$$\frac{\pi}{\sin^2 \frac{\pi}{x}}=\frac{\pi}{\sin \frac{\pi}{x}\sin \frac{\pi}{x}}=\frac{1}{\sin \frac{\pi}{x} / \cos \frac{\pi}{x}}\frac{\pi}{\cos\frac{\pi}{x}\sin \frac{\pi}{x}}=\frac{1}{\tan \frac{\pi}{x}}\frac{\pi}{\cos\frac{\pi}{x}\sin \frac{\pi}{x}}$$ $$=\frac{x}{\tan \frac{\pi}{x}}\frac{\pi}{x\cos\frac{\pi}{x}\sin \frac{\pi}{x}}=\frac{x}{\tan \frac{\pi}{x}}\frac{\pi/x}{\cos\frac{\pi}{x}\sin \frac{\pi}{x}}.$$

$$\frac{\pi}{\sin^2 \frac{\pi}{x}}=\frac{x}{\tan \frac{\pi}{x}}\frac{\pi/x}{\cos\frac{\pi}{x}\sin \frac{\pi}{x}}.$$

Now, you need to estimate the second factor.

user · Answer 4 · 2020-09-01T17:31:31.683

We can cancel out both sides a factor $\sin \frac{\pi}{x}>0$ to obtain

$$\frac{\pi}{\sin^2 \frac{\pi}{x}} \gt \frac{x}{\tan \frac{\pi}{x}}\iff \frac{\pi}{\sin^2 \frac{\pi}{x}} \gt \frac{x}{\frac{\sin\frac{\pi}{x}}{\cos\frac{\pi}{x}}}$$

$$\iff \frac{\pi}{\sin \frac{\pi}{x}} \gt \frac{x}{\frac{1}{\cos\frac{\pi}{x}}} \iff \frac \pi x> \sin \frac{\pi}{x}\cos \frac{\pi}{x}=\frac12 \sin \frac{2\pi}{x}$$

$$\iff \sin \frac{2\pi}{x}< \frac{2\pi}{x}$$

a well known inequality which holds for $\frac{2\pi}{x}>0$ and therefore for $x>2$.

Refer to the related

Something wrong with that proof? Thanks for any advice! – user Sep 01 '20 at 17:42 — user, Sep 01 '20 at 17:42

Prove $\frac{\pi}{\sin^2 \frac{\pi}{x}} \gt \frac{x}{\tan \frac{\pi}{x}}$

4 Answers4