454. Mathematical note: "trig. ineg."- GMB 6/1992, page 201.
by Virgil Nicula, May 17, 2017, 8:04 AM
P0. Ascertain 
where ![$f:\left(0,\frac {\pi}2\right]\rightarrow \mathbb R\ ,\ f(x)=\frac {\sin x-x\cos x}{x-\sin x}$](//latex.artofproblemsolving.com/b/0/f/b0fc5111c33d1c5cce113cf1ae56634089a91f95.png)
Proof.
and 
If we'll prove that the function continuous 
is strict monotonous, evidently strict decreasing
because
and 
then from the Darboux's property get 
Indeed, 
![$\sin x\cdot\left[x-\frac {1-\cos x}{2\sin x}\cdot\left(1-\sqrt{4\cos x+5}\right)\right]\cdot\left[x-\frac {1-\cos x}{2\sin x}\cdot \left(1+\sqrt{4\cos x+5}\right)\right]\implies$](//latex.artofproblemsolving.com/1/2/7/127c85401e1fee1efc888f4e97ed4291ea2ff499.png)
But ![$g'(x)\ \mathrm{.a.s.}\ (1-\cos x)\left[(1+2\cos x\sqrt{4\cos x+5}-\left(2\cos^2x+4\cos x+3\right)\right]\ \mathrm{.a.s.}\ (1+2\cos x)^2(4\cos x+5)$](//latex.artofproblemsolving.com/1/c/c/1ccb5c5314b468c703d21ceda6f9f704bf5f8645.png)
In conclusion,
for any 
the function is decreasing and 
i.e. ![$\boxed{(\forall ) x\in\left(0,\frac {\pi}2\right]\ ,\ \frac 2{\pi -2}<\frac {\sin x-x\cos x}{x-\sin x}<2}\ (*)\ .$](//latex.artofproblemsolving.com/2/1/5/2156dc25e5cb79f5d2fd60936cbca0dafda0b495.png)
Remark 1. I'll write the inequality
thus: 
The inequality 
is equivalent with the inequality and is stronger than the wellknown inequality 
Remark 2. I used the equivalence relation over
Geometrical interpretation. I'll look for the most far point
to the left side of the axis 
where 
so that for any point 
where 
the length 
of the arc ![$l(\overarc{AM})<[AT]\ ,$](//latex.artofproblemsolving.com/a/b/f/abf34fa37e2a9ed139cea1928a616c7bb0e19533.png)
where 
and 
For 
obtain 
for any 
Let 
i.e. 
and 
From 
obtain 
Therefore, 
for any 
i.e. 
With other words, the most far point 
to the left side of the axis 
for which 
is 
where 
Applications.
.
where ![$f:\left(0,\frac {\pi}2\right]\rightarrow \mathbb R\ ,\ f(x)=\frac {\sin x-x\cos x}{x-\sin x}$](http://latex.artofproblemsolving.com/b/0/f/b0fc5111c33d1c5cce113cf1ae56634089a91f95.png)
Proof.
and 
If we'll prove that the function continuous 
is strict monotonous, evidently strict decreasingbecause
and 
then from the Darboux's property get 
Indeed, 
![$\sin x\cdot\left[x-\frac {1-\cos x}{2\sin x}\cdot\left(1-\sqrt{4\cos x+5}\right)\right]\cdot\left[x-\frac {1-\cos x}{2\sin x}\cdot \left(1+\sqrt{4\cos x+5}\right)\right]\implies$](http://latex.artofproblemsolving.com/1/2/7/127c85401e1fee1efc888f4e97ed4291ea2ff499.png)
But ![$g'(x)\ \mathrm{.a.s.}\ (1-\cos x)\left[(1+2\cos x\sqrt{4\cos x+5}-\left(2\cos^2x+4\cos x+3\right)\right]\ \mathrm{.a.s.}\ (1+2\cos x)^2(4\cos x+5)$](http://latex.artofproblemsolving.com/1/c/c/1ccb5c5314b468c703d21ceda6f9f704bf5f8645.png)
In conclusion,
for any 
the function is decreasing and 
i.e. ![$\boxed{(\forall ) x\in\left(0,\frac {\pi}2\right]\ ,\ \frac 2{\pi -2}<\frac {\sin x-x\cos x}{x-\sin x}<2}\ (*)\ .$](http://latex.artofproblemsolving.com/2/1/5/2156dc25e5cb79f5d2fd60936cbca0dafda0b495.png)
Remark 1. I'll write the inequality
thus: 
The inequality 
is equivalent with the inequality
and is stronger than the wellknown inequality 
Remark 2. I used the equivalence relation over
Geometrical interpretation. I'll look for the most far point
to the left side of the axis 
where 
so that for any point 
where 
the length 
of the arc ![$l(\overarc{AM})<[AT]\ ,$](http://latex.artofproblemsolving.com/a/b/f/abf34fa37e2a9ed139cea1928a616c7bb0e19533.png)
where 
and 
For 
obtain 
for any 
Let 
i.e. 
and 
From 
obtain 
Therefore, 
for any 
i.e. 
With other words, the most far point 
to the left side of the axis 
for which 
is 
where 
Applications.
.
This post has been edited 68 times. Last edited by Virgil Nicula, Jul 3, 2017, 8:10 AM
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