Using Mean Value Theorem, Prove that ${tan xover x}>{xoversin x}$












4












$begingroup$


Prove using Mean Value Theorem,




$${tan xover x}>{xoversin x} spaceforall space x spaceepsilon (0, pi/2) $$




Attempt::



$f(x) = x-sin x$



$f'(x) = 1-cos x > 0 $



Hence $x-sin x>0, {xoversin x}>1$



Similarly, I got ${tan xover x}>1$



But how do I compare them and get the required inequality?






calculus limits inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $xin mathbb R$ or in some interval? (In $mathbb R$ it does not hold)
    $endgroup$
    – Jimmy R.
    Dec 2 '16 at 14:45












  • $begingroup$
    Isn't $x$ restricted to any interval?
    $endgroup$
    – ajotatxe
    Dec 2 '16 at 14:45






  • 1




    $begingroup$
    Prove $tanx/x >1$ and $sinx/x<1$ and any interval restriction ?
    $endgroup$
    – papabiceps
    Dec 2 '16 at 14:45








  • 1




    $begingroup$
    Interval is between 0<x<pi/2
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:05
















4












$begingroup$


Prove using Mean Value Theorem,




$${tan xover x}>{xoversin x} spaceforall space x spaceepsilon (0, pi/2) $$




Attempt::



$f(x) = x-sin x$



$f'(x) = 1-cos x > 0 $



Hence $x-sin x>0, {xoversin x}>1$



Similarly, I got ${tan xover x}>1$



But how do I compare them and get the required inequality?






calculus limits inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $xin mathbb R$ or in some interval? (In $mathbb R$ it does not hold)
    $endgroup$
    – Jimmy R.
    Dec 2 '16 at 14:45












  • $begingroup$
    Isn't $x$ restricted to any interval?
    $endgroup$
    – ajotatxe
    Dec 2 '16 at 14:45






  • 1




    $begingroup$
    Prove $tanx/x >1$ and $sinx/x<1$ and any interval restriction ?
    $endgroup$
    – papabiceps
    Dec 2 '16 at 14:45








  • 1




    $begingroup$
    Interval is between 0<x<pi/2
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:05














4












4








4





$begingroup$


Prove using Mean Value Theorem,




$${tan xover x}>{xoversin x} spaceforall space x spaceepsilon (0, pi/2) $$




Attempt::



$f(x) = x-sin x$



$f'(x) = 1-cos x > 0 $



Hence $x-sin x>0, {xoversin x}>1$



Similarly, I got ${tan xover x}>1$



But how do I compare them and get the required inequality?






calculus limits inequality







share|cite|improve this question











$endgroup$




Prove using Mean Value Theorem,




$${tan xover x}>{xoversin x} spaceforall space x spaceepsilon (0, pi/2) $$




Attempt::



$f(x) = x-sin x$



$f'(x) = 1-cos x > 0 $



Hence $x-sin x>0, {xoversin x}>1$



Similarly, I got ${tan xover x}>1$



But how do I compare them and get the required inequality?






calculus limits inequality




calculus limits inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 5:58









tatan

5,79762759




5,79762759










asked Dec 2 '16 at 14:41









user3442005user3442005

305




305












  • $begingroup$
    For $xin mathbb R$ or in some interval? (In $mathbb R$ it does not hold)
    $endgroup$
    – Jimmy R.
    Dec 2 '16 at 14:45












  • $begingroup$
    Isn't $x$ restricted to any interval?
    $endgroup$
    – ajotatxe
    Dec 2 '16 at 14:45






  • 1




    $begingroup$
    Prove $tanx/x >1$ and $sinx/x<1$ and any interval restriction ?
    $endgroup$
    – papabiceps
    Dec 2 '16 at 14:45








  • 1




    $begingroup$
    Interval is between 0<x<pi/2
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:05


















  • $begingroup$
    For $xin mathbb R$ or in some interval? (In $mathbb R$ it does not hold)
    $endgroup$
    – Jimmy R.
    Dec 2 '16 at 14:45












  • $begingroup$
    Isn't $x$ restricted to any interval?
    $endgroup$
    – ajotatxe
    Dec 2 '16 at 14:45






  • 1




    $begingroup$
    Prove $tanx/x >1$ and $sinx/x<1$ and any interval restriction ?
    $endgroup$
    – papabiceps
    Dec 2 '16 at 14:45








  • 1




    $begingroup$
    Interval is between 0<x<pi/2
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:05
















$begingroup$
For $xin mathbb R$ or in some interval? (In $mathbb R$ it does not hold)
$endgroup$
– Jimmy R.
Dec 2 '16 at 14:45






$begingroup$
For $xin mathbb R$ or in some interval? (In $mathbb R$ it does not hold)
$endgroup$
– Jimmy R.
Dec 2 '16 at 14:45














$begingroup$
Isn't $x$ restricted to any interval?
$endgroup$
– ajotatxe
Dec 2 '16 at 14:45




$begingroup$
Isn't $x$ restricted to any interval?
$endgroup$
– ajotatxe
Dec 2 '16 at 14:45




1




1




$begingroup$
Prove $tanx/x >1$ and $sinx/x<1$ and any interval restriction ?
$endgroup$
– papabiceps
Dec 2 '16 at 14:45






$begingroup$
Prove $tanx/x >1$ and $sinx/x<1$ and any interval restriction ?
$endgroup$
– papabiceps
Dec 2 '16 at 14:45






1




1




$begingroup$
Interval is between 0<x<pi/2
$endgroup$
– user3442005
Dec 2 '16 at 15:05




$begingroup$
Interval is between 0<x<pi/2
$endgroup$
– user3442005
Dec 2 '16 at 15:05










2 Answers
2






active

oldest

votes


















1












$begingroup$

We need to prove that $f(x)>0$, where $f(x)=frac{sin{x}}{sqrt{cos{x}}}-x$.



But $f'(x)=frac{cos xsqrt{cos x}+frac{sin^2x}{2sqrt{cos x}}}{cos{x}}-1=frac{(sqrt{cos^3{x}}-1)^2+cos^2x(1-cos{x})}{2sqrt{cos^3x}}>0$.



Thus, $f(x)>f(0)=0$.



Done!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is f(x) = sinx/root(cosx) -x ?That isnt the question.
    $endgroup$
    – user3442005
    Dec 2 '16 at 16:58










  • $begingroup$
    @user3442005 We need to prove that $frac{sin^2x}{cos{x}}>x^2$, which is $f(x)>0$.
    $endgroup$
    – Michael Rozenberg
    Dec 2 '16 at 17:27










  • $begingroup$
    Where did you use Mean Value Theorem?
    $endgroup$
    – user261263
    Dec 3 '16 at 6:05





















1












$begingroup$

For each $x in (0,2pi),$ the MVT(mean value theorem) makes sure that there exists $c_x$ with $0 < c_x < x$ such that



$$frac{sin x}{x} = cos c_x < 1.$$



With $f(x) = sin x$ we have $sin x = sin x - sin 0 = f'(c_x)(x-0)= (cos c_x)x$



where $0 < c_x < x$.



Now for $displaystyle frac{tan x}{x}>1$



$f(x) = tan x$ then $f'(x) = sec^2(x)$. So $sec^2(x)$ = $displaystyle frac{tan x}{x}$



For each $x in (0,2pi)$ $sec^2(x) >1$. So $displaystyle frac{tan x}{x}$ $>1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I had forgotten to mention the limits, it is from (0, pi/2)
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:13










  • $begingroup$
    Then my answer holds true for your given limits.
    $endgroup$
    – papabiceps
    Dec 2 '16 at 15:15










  • $begingroup$
    According to the solution, x/sinx>1 and tanx/x>1. How do I compare the two inequalities and get tanx/x > x/sinx?
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:24










  • $begingroup$
    The comparison is with x/sin x, not sin x/x, so this does not solve the problem.
    $endgroup$
    – marty cohen
    Dec 2 '16 at 15:30










  • $begingroup$
    $tan$ function is not well defined on $x in (0,2pi)$
    $endgroup$
    – user261263
    Dec 3 '16 at 6:04











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

We need to prove that $f(x)>0$, where $f(x)=frac{sin{x}}{sqrt{cos{x}}}-x$.



But $f'(x)=frac{cos xsqrt{cos x}+frac{sin^2x}{2sqrt{cos x}}}{cos{x}}-1=frac{(sqrt{cos^3{x}}-1)^2+cos^2x(1-cos{x})}{2sqrt{cos^3x}}>0$.



Thus, $f(x)>f(0)=0$.



Done!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is f(x) = sinx/root(cosx) -x ?That isnt the question.
    $endgroup$
    – user3442005
    Dec 2 '16 at 16:58










  • $begingroup$
    @user3442005 We need to prove that $frac{sin^2x}{cos{x}}>x^2$, which is $f(x)>0$.
    $endgroup$
    – Michael Rozenberg
    Dec 2 '16 at 17:27










  • $begingroup$
    Where did you use Mean Value Theorem?
    $endgroup$
    – user261263
    Dec 3 '16 at 6:05


















1












$begingroup$

We need to prove that $f(x)>0$, where $f(x)=frac{sin{x}}{sqrt{cos{x}}}-x$.



But $f'(x)=frac{cos xsqrt{cos x}+frac{sin^2x}{2sqrt{cos x}}}{cos{x}}-1=frac{(sqrt{cos^3{x}}-1)^2+cos^2x(1-cos{x})}{2sqrt{cos^3x}}>0$.



Thus, $f(x)>f(0)=0$.



Done!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is f(x) = sinx/root(cosx) -x ?That isnt the question.
    $endgroup$
    – user3442005
    Dec 2 '16 at 16:58










  • $begingroup$
    @user3442005 We need to prove that $frac{sin^2x}{cos{x}}>x^2$, which is $f(x)>0$.
    $endgroup$
    – Michael Rozenberg
    Dec 2 '16 at 17:27










  • $begingroup$
    Where did you use Mean Value Theorem?
    $endgroup$
    – user261263
    Dec 3 '16 at 6:05
















1












1








1





$begingroup$

We need to prove that $f(x)>0$, where $f(x)=frac{sin{x}}{sqrt{cos{x}}}-x$.



But $f'(x)=frac{cos xsqrt{cos x}+frac{sin^2x}{2sqrt{cos x}}}{cos{x}}-1=frac{(sqrt{cos^3{x}}-1)^2+cos^2x(1-cos{x})}{2sqrt{cos^3x}}>0$.



Thus, $f(x)>f(0)=0$.



Done!






share|cite|improve this answer









$endgroup$



We need to prove that $f(x)>0$, where $f(x)=frac{sin{x}}{sqrt{cos{x}}}-x$.



But $f'(x)=frac{cos xsqrt{cos x}+frac{sin^2x}{2sqrt{cos x}}}{cos{x}}-1=frac{(sqrt{cos^3{x}}-1)^2+cos^2x(1-cos{x})}{2sqrt{cos^3x}}>0$.



Thus, $f(x)>f(0)=0$.



Done!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '16 at 16:44









Michael RozenbergMichael Rozenberg

105k1892198




105k1892198












  • $begingroup$
    How is f(x) = sinx/root(cosx) -x ?That isnt the question.
    $endgroup$
    – user3442005
    Dec 2 '16 at 16:58










  • $begingroup$
    @user3442005 We need to prove that $frac{sin^2x}{cos{x}}>x^2$, which is $f(x)>0$.
    $endgroup$
    – Michael Rozenberg
    Dec 2 '16 at 17:27










  • $begingroup$
    Where did you use Mean Value Theorem?
    $endgroup$
    – user261263
    Dec 3 '16 at 6:05




















  • $begingroup$
    How is f(x) = sinx/root(cosx) -x ?That isnt the question.
    $endgroup$
    – user3442005
    Dec 2 '16 at 16:58










  • $begingroup$
    @user3442005 We need to prove that $frac{sin^2x}{cos{x}}>x^2$, which is $f(x)>0$.
    $endgroup$
    – Michael Rozenberg
    Dec 2 '16 at 17:27










  • $begingroup$
    Where did you use Mean Value Theorem?
    $endgroup$
    – user261263
    Dec 3 '16 at 6:05


















$begingroup$
How is f(x) = sinx/root(cosx) -x ?That isnt the question.
$endgroup$
– user3442005
Dec 2 '16 at 16:58




$begingroup$
How is f(x) = sinx/root(cosx) -x ?That isnt the question.
$endgroup$
– user3442005
Dec 2 '16 at 16:58












$begingroup$
@user3442005 We need to prove that $frac{sin^2x}{cos{x}}>x^2$, which is $f(x)>0$.
$endgroup$
– Michael Rozenberg
Dec 2 '16 at 17:27




$begingroup$
@user3442005 We need to prove that $frac{sin^2x}{cos{x}}>x^2$, which is $f(x)>0$.
$endgroup$
– Michael Rozenberg
Dec 2 '16 at 17:27












$begingroup$
Where did you use Mean Value Theorem?
$endgroup$
– user261263
Dec 3 '16 at 6:05






$begingroup$
Where did you use Mean Value Theorem?
$endgroup$
– user261263
Dec 3 '16 at 6:05













1












$begingroup$

For each $x in (0,2pi),$ the MVT(mean value theorem) makes sure that there exists $c_x$ with $0 < c_x < x$ such that



$$frac{sin x}{x} = cos c_x < 1.$$



With $f(x) = sin x$ we have $sin x = sin x - sin 0 = f'(c_x)(x-0)= (cos c_x)x$



where $0 < c_x < x$.



Now for $displaystyle frac{tan x}{x}>1$



$f(x) = tan x$ then $f'(x) = sec^2(x)$. So $sec^2(x)$ = $displaystyle frac{tan x}{x}$



For each $x in (0,2pi)$ $sec^2(x) >1$. So $displaystyle frac{tan x}{x}$ $>1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I had forgotten to mention the limits, it is from (0, pi/2)
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:13










  • $begingroup$
    Then my answer holds true for your given limits.
    $endgroup$
    – papabiceps
    Dec 2 '16 at 15:15










  • $begingroup$
    According to the solution, x/sinx>1 and tanx/x>1. How do I compare the two inequalities and get tanx/x > x/sinx?
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:24










  • $begingroup$
    The comparison is with x/sin x, not sin x/x, so this does not solve the problem.
    $endgroup$
    – marty cohen
    Dec 2 '16 at 15:30










  • $begingroup$
    $tan$ function is not well defined on $x in (0,2pi)$
    $endgroup$
    – user261263
    Dec 3 '16 at 6:04
















1












$begingroup$

For each $x in (0,2pi),$ the MVT(mean value theorem) makes sure that there exists $c_x$ with $0 < c_x < x$ such that



$$frac{sin x}{x} = cos c_x < 1.$$



With $f(x) = sin x$ we have $sin x = sin x - sin 0 = f'(c_x)(x-0)= (cos c_x)x$



where $0 < c_x < x$.



Now for $displaystyle frac{tan x}{x}>1$



$f(x) = tan x$ then $f'(x) = sec^2(x)$. So $sec^2(x)$ = $displaystyle frac{tan x}{x}$



For each $x in (0,2pi)$ $sec^2(x) >1$. So $displaystyle frac{tan x}{x}$ $>1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I had forgotten to mention the limits, it is from (0, pi/2)
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:13










  • $begingroup$
    Then my answer holds true for your given limits.
    $endgroup$
    – papabiceps
    Dec 2 '16 at 15:15










  • $begingroup$
    According to the solution, x/sinx>1 and tanx/x>1. How do I compare the two inequalities and get tanx/x > x/sinx?
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:24










  • $begingroup$
    The comparison is with x/sin x, not sin x/x, so this does not solve the problem.
    $endgroup$
    – marty cohen
    Dec 2 '16 at 15:30










  • $begingroup$
    $tan$ function is not well defined on $x in (0,2pi)$
    $endgroup$
    – user261263
    Dec 3 '16 at 6:04














1












1








1





$begingroup$

For each $x in (0,2pi),$ the MVT(mean value theorem) makes sure that there exists $c_x$ with $0 < c_x < x$ such that



$$frac{sin x}{x} = cos c_x < 1.$$



With $f(x) = sin x$ we have $sin x = sin x - sin 0 = f'(c_x)(x-0)= (cos c_x)x$



where $0 < c_x < x$.



Now for $displaystyle frac{tan x}{x}>1$



$f(x) = tan x$ then $f'(x) = sec^2(x)$. So $sec^2(x)$ = $displaystyle frac{tan x}{x}$



For each $x in (0,2pi)$ $sec^2(x) >1$. So $displaystyle frac{tan x}{x}$ $>1$.






share|cite|improve this answer











$endgroup$



For each $x in (0,2pi),$ the MVT(mean value theorem) makes sure that there exists $c_x$ with $0 < c_x < x$ such that



$$frac{sin x}{x} = cos c_x < 1.$$



With $f(x) = sin x$ we have $sin x = sin x - sin 0 = f'(c_x)(x-0)= (cos c_x)x$



where $0 < c_x < x$.



Now for $displaystyle frac{tan x}{x}>1$



$f(x) = tan x$ then $f'(x) = sec^2(x)$. So $sec^2(x)$ = $displaystyle frac{tan x}{x}$



For each $x in (0,2pi)$ $sec^2(x) >1$. So $displaystyle frac{tan x}{x}$ $>1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 '16 at 15:04









juantheron

34.3k1148143




34.3k1148143










answered Dec 2 '16 at 14:55









papabicepspapabiceps

322113




322113












  • $begingroup$
    I had forgotten to mention the limits, it is from (0, pi/2)
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:13










  • $begingroup$
    Then my answer holds true for your given limits.
    $endgroup$
    – papabiceps
    Dec 2 '16 at 15:15










  • $begingroup$
    According to the solution, x/sinx>1 and tanx/x>1. How do I compare the two inequalities and get tanx/x > x/sinx?
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:24










  • $begingroup$
    The comparison is with x/sin x, not sin x/x, so this does not solve the problem.
    $endgroup$
    – marty cohen
    Dec 2 '16 at 15:30










  • $begingroup$
    $tan$ function is not well defined on $x in (0,2pi)$
    $endgroup$
    – user261263
    Dec 3 '16 at 6:04


















  • $begingroup$
    I had forgotten to mention the limits, it is from (0, pi/2)
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:13










  • $begingroup$
    Then my answer holds true for your given limits.
    $endgroup$
    – papabiceps
    Dec 2 '16 at 15:15










  • $begingroup$
    According to the solution, x/sinx>1 and tanx/x>1. How do I compare the two inequalities and get tanx/x > x/sinx?
    $endgroup$
    – user3442005
    Dec 2 '16 at 15:24










  • $begingroup$
    The comparison is with x/sin x, not sin x/x, so this does not solve the problem.
    $endgroup$
    – marty cohen
    Dec 2 '16 at 15:30










  • $begingroup$
    $tan$ function is not well defined on $x in (0,2pi)$
    $endgroup$
    – user261263
    Dec 3 '16 at 6:04
















$begingroup$
I had forgotten to mention the limits, it is from (0, pi/2)
$endgroup$
– user3442005
Dec 2 '16 at 15:13




$begingroup$
I had forgotten to mention the limits, it is from (0, pi/2)
$endgroup$
– user3442005
Dec 2 '16 at 15:13












$begingroup$
Then my answer holds true for your given limits.
$endgroup$
– papabiceps
Dec 2 '16 at 15:15




$begingroup$
Then my answer holds true for your given limits.
$endgroup$
– papabiceps
Dec 2 '16 at 15:15












$begingroup$
According to the solution, x/sinx>1 and tanx/x>1. How do I compare the two inequalities and get tanx/x > x/sinx?
$endgroup$
– user3442005
Dec 2 '16 at 15:24




$begingroup$
According to the solution, x/sinx>1 and tanx/x>1. How do I compare the two inequalities and get tanx/x > x/sinx?
$endgroup$
– user3442005
Dec 2 '16 at 15:24












$begingroup$
The comparison is with x/sin x, not sin x/x, so this does not solve the problem.
$endgroup$
– marty cohen
Dec 2 '16 at 15:30




$begingroup$
The comparison is with x/sin x, not sin x/x, so this does not solve the problem.
$endgroup$
– marty cohen
Dec 2 '16 at 15:30












$begingroup$
$tan$ function is not well defined on $x in (0,2pi)$
$endgroup$
– user261263
Dec 3 '16 at 6:04




$begingroup$
$tan$ function is not well defined on $x in (0,2pi)$
$endgroup$
– user261263
Dec 3 '16 at 6:04


















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