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Prove that, $cos xge1-2xoverpi forall xin[0,piover2]$

0

1

$begingroup$

By drawing the graph of $y=cos x$ and $y=1-2xoverpi$ on $[0,piover2]$, it looks obvious due to convex nature of cosine function, which follows-


But how to prove it rigorously, using Taylor Series expansion or integration or by any other analysis.

Can anybody prove it?

Thanks for assistance in advance.

real-analysis trigonometry inequality

share|cite|improve this question

asked Mar 31 at 14:59

Biswarup SahaBiswarup Saha

656110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take derivative and use $sin x < x$ in that interval
  $endgroup$
  – J. W. Tanner
  Mar 31 at 15:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just to add to what J. W. Said , take the derivative of $cos(x)-1+2/pi x$
  $endgroup$
  – Sorfosh
  Mar 31 at 15:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Sorfosh do you want to write $2x/pi$?
  $endgroup$
  – Biswarup Saha
  Mar 31 at 15:25
  
  

  
  
  
  

add a comment | 

0

1

$begingroup$

By drawing the graph of $y=cos x$ and $y=1-2xoverpi$ on $[0,piover2]$, it looks obvious due to convex nature of cosine function, which follows-


But how to prove it rigorously, using Taylor Series expansion or integration or by any other analysis.

Can anybody prove it?

Thanks for assistance in advance.

real-analysis trigonometry inequality

share|cite|improve this question

asked Mar 31 at 14:59

Biswarup SahaBiswarup Saha

656110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take derivative and use $sin x < x$ in that interval
  $endgroup$
  – J. W. Tanner
  Mar 31 at 15:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just to add to what J. W. Said , take the derivative of $cos(x)-1+2/pi x$
  $endgroup$
  – Sorfosh
  Mar 31 at 15:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Sorfosh do you want to write $2x/pi$?
  $endgroup$
  – Biswarup Saha
  Mar 31 at 15:25
  
  

  
  
  
  

add a comment | 

$begingroup$

By drawing the graph of $y=cos x$ and $y=1-2xoverpi$ on $[0,piover2]$, it looks obvious due to convex nature of cosine function, which follows-


But how to prove it rigorously, using Taylor Series expansion or integration or by any other analysis.

Can anybody prove it?

Thanks for assistance in advance.

real-analysis trigonometry inequality

share|cite|improve this question

asked Mar 31 at 14:59

Biswarup SahaBiswarup Saha

656110

$endgroup$

share|cite|improve this question

asked Mar 31 at 14:59

Biswarup SahaBiswarup Saha

656110

share|cite|improve this question

asked Mar 31 at 14:59

Biswarup SahaBiswarup Saha

656110

asked Mar 31 at 14:59

Biswarup SahaBiswarup Saha

656110

asked Mar 31 at 14:59

Biswarup SahaBiswarup Saha

656110

2 Answers
2

active

oldest

votes

3

$begingroup$

Let $f(x)=cos(x)+dfrac2pi x$. Then $f'(x)=dfrac2pi-sin(x)$ and, since $dfrac2piin(0,1)$ and $sin$ is increasing in $left[0,fracpi2right]$, there is one and only one $x_0inleft[0,fracpi2right]$ such that $sin(x_0)=dfrac2pi$. So, $f$ is increasing on$[0,x_0]$ and decreasing on $left[x_0,fracpi2right]$. But then, since $f(0)=fleft(fracpi2right)=1$, you have $left(forall xinleft[0,fracpi2right]right):f(x)geqslant1$, which is exactly what you want to prove.

share|cite|improve this answer

edited Mar 31 at 17:18

answered Mar 31 at 15:05

José Carlos SantosJosé Carlos Santos

174k23134243

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think there is some mistake in your function
  $endgroup$
  – Biswarup Saha
  Mar 31 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited my answer. What do you think now?
  $endgroup$
  – José Carlos Santos
  Mar 31 at 17:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) I wrote up an answer that was close to yours before I saw yours.
  $endgroup$
  – robjohn♦
  Mar 31 at 17:45
  

  
  
  
  

add a comment | 

1

$begingroup$

Simple: the function $cos x$ is concave on the interval $[0,pi/2]$, hence any chord on this interval is under the arc it subtends. In particular, the chord joining the $x$-intercept $(pi/2,0)$ and the $y$-intercept $(0,1)$.

Now, an equation of the line through $(a,0)$ and $(0,b)$ is $dfrac xa+dfrac yb=1$, so here we have the equation
$$frac2xpi +frac y1=1iff y=1-frac2xpi,$$
and the chord being under the curve arc translates as
$$1-frac2xpi lecos x.$$

share|cite|improve this answer

edited Mar 31 at 15:45

answered Mar 31 at 15:17

BernardBernard

124k741117

$endgroup$

add a comment | 

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3

$begingroup$

Let $f(x)=cos(x)+dfrac2pi x$. Then $f'(x)=dfrac2pi-sin(x)$ and, since $dfrac2piin(0,1)$ and $sin$ is increasing in $left[0,fracpi2right]$, there is one and only one $x_0inleft[0,fracpi2right]$ such that $sin(x_0)=dfrac2pi$. So, $f$ is increasing on$[0,x_0]$ and decreasing on $left[x_0,fracpi2right]$. But then, since $f(0)=fleft(fracpi2right)=1$, you have $left(forall xinleft[0,fracpi2right]right):f(x)geqslant1$, which is exactly what you want to prove.

share|cite|improve this answer

edited Mar 31 at 17:18

answered Mar 31 at 15:05

José Carlos SantosJosé Carlos Santos

174k23134243

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think there is some mistake in your function
  $endgroup$
  – Biswarup Saha
  Mar 31 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited my answer. What do you think now?
  $endgroup$
  – José Carlos Santos
  Mar 31 at 17:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) I wrote up an answer that was close to yours before I saw yours.
  $endgroup$
  – robjohn♦
  Mar 31 at 17:45
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $f(x)=cos(x)+dfrac2pi x$. Then $f'(x)=dfrac2pi-sin(x)$ and, since $dfrac2piin(0,1)$ and $sin$ is increasing in $left[0,fracpi2right]$, there is one and only one $x_0inleft[0,fracpi2right]$ such that $sin(x_0)=dfrac2pi$. So, $f$ is increasing on$[0,x_0]$ and decreasing on $left[x_0,fracpi2right]$. But then, since $f(0)=fleft(fracpi2right)=1$, you have $left(forall xinleft[0,fracpi2right]right):f(x)geqslant1$, which is exactly what you want to prove.

share|cite|improve this answer

edited Mar 31 at 17:18

answered Mar 31 at 15:05

José Carlos SantosJosé Carlos Santos

174k23134243

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think there is some mistake in your function
  $endgroup$
  – Biswarup Saha
  Mar 31 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited my answer. What do you think now?
  $endgroup$
  – José Carlos Santos
  Mar 31 at 17:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) I wrote up an answer that was close to yours before I saw yours.
  $endgroup$
  – robjohn♦
  Mar 31 at 17:45
  

  
  
  
  

add a comment | 

$begingroup$

Let $f(x)=cos(x)+dfrac2pi x$. Then $f'(x)=dfrac2pi-sin(x)$ and, since $dfrac2piin(0,1)$ and $sin$ is increasing in $left[0,fracpi2right]$, there is one and only one $x_0inleft[0,fracpi2right]$ such that $sin(x_0)=dfrac2pi$. So, $f$ is increasing on$[0,x_0]$ and decreasing on $left[x_0,fracpi2right]$. But then, since $f(0)=fleft(fracpi2right)=1$, you have $left(forall xinleft[0,fracpi2right]right):f(x)geqslant1$, which is exactly what you want to prove.

share|cite|improve this answer

edited Mar 31 at 17:18

answered Mar 31 at 15:05

José Carlos SantosJosé Carlos Santos

174k23134243

$endgroup$

share|cite|improve this answer

edited Mar 31 at 17:18

answered Mar 31 at 15:05

José Carlos SantosJosé Carlos Santos

174k23134243

answered Mar 31 at 15:05

José Carlos SantosJosé Carlos Santos

174k23134243

answered Mar 31 at 15:05

José Carlos SantosJosé Carlos Santos

174k23134243

1

$begingroup$

Simple: the function $cos x$ is concave on the interval $[0,pi/2]$, hence any chord on this interval is under the arc it subtends. In particular, the chord joining the $x$-intercept $(pi/2,0)$ and the $y$-intercept $(0,1)$.

Now, an equation of the line through $(a,0)$ and $(0,b)$ is $dfrac xa+dfrac yb=1$, so here we have the equation
$$frac2xpi +frac y1=1iff y=1-frac2xpi,$$
and the chord being under the curve arc translates as
$$1-frac2xpi lecos x.$$

share|cite|improve this answer

edited Mar 31 at 15:45

answered Mar 31 at 15:17

BernardBernard

124k741117

$endgroup$

add a comment | 

1

$begingroup$

Simple: the function $cos x$ is concave on the interval $[0,pi/2]$, hence any chord on this interval is under the arc it subtends. In particular, the chord joining the $x$-intercept $(pi/2,0)$ and the $y$-intercept $(0,1)$.

Now, an equation of the line through $(a,0)$ and $(0,b)$ is $dfrac xa+dfrac yb=1$, so here we have the equation
$$frac2xpi +frac y1=1iff y=1-frac2xpi,$$
and the chord being under the curve arc translates as
$$1-frac2xpi lecos x.$$

share|cite|improve this answer

edited Mar 31 at 15:45

answered Mar 31 at 15:17

BernardBernard

124k741117

$endgroup$

add a comment | 

$begingroup$

Simple: the function $cos x$ is concave on the interval $[0,pi/2]$, hence any chord on this interval is under the arc it subtends. In particular, the chord joining the $x$-intercept $(pi/2,0)$ and the $y$-intercept $(0,1)$.

Now, an equation of the line through $(a,0)$ and $(0,b)$ is $dfrac xa+dfrac yb=1$, so here we have the equation
$$frac2xpi +frac y1=1iff y=1-frac2xpi,$$
and the chord being under the curve arc translates as
$$1-frac2xpi lecos x.$$

share|cite|improve this answer

edited Mar 31 at 15:45

answered Mar 31 at 15:17

BernardBernard

124k741117

$endgroup$

share|cite|improve this answer

edited Mar 31 at 15:45

answered Mar 31 at 15:17

BernardBernard

124k741117

answered Mar 31 at 15:17

BernardBernard

124k741117

answered Mar 31 at 15:17

BernardBernard

124k741117