Prove $sin x$ is uniformly continuous on $mathbb R$












18












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How do I prove $sin x$ is uniformly continuous on $mathbb R$ with delta and epsilon?



I proved geometrically that $sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|sin x_1 - sin x_2|le|sin x_1|+|sin x_2|<|x_1|+|x_2|$$



But this doesn't help me much finding a delta...



Thanks for any help!



P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).






calculus functions continuity







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  • 1




    $begingroup$
    I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
    $endgroup$
    – Michael Hardy
    Jan 1 '13 at 19:35












  • $begingroup$
    This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
    $endgroup$
    – Julien
    Jan 1 '13 at 19:41


















18












$begingroup$


How do I prove $sin x$ is uniformly continuous on $mathbb R$ with delta and epsilon?



I proved geometrically that $sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|sin x_1 - sin x_2|le|sin x_1|+|sin x_2|<|x_1|+|x_2|$$



But this doesn't help me much finding a delta...



Thanks for any help!



P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).






calculus functions continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
    $endgroup$
    – Michael Hardy
    Jan 1 '13 at 19:35












  • $begingroup$
    This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
    $endgroup$
    – Julien
    Jan 1 '13 at 19:41
















18












18








18


11



$begingroup$


How do I prove $sin x$ is uniformly continuous on $mathbb R$ with delta and epsilon?



I proved geometrically that $sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|sin x_1 - sin x_2|le|sin x_1|+|sin x_2|<|x_1|+|x_2|$$



But this doesn't help me much finding a delta...



Thanks for any help!



P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).






calculus functions continuity







share|cite|improve this question











$endgroup$




How do I prove $sin x$ is uniformly continuous on $mathbb R$ with delta and epsilon?



I proved geometrically that $sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|sin x_1 - sin x_2|le|sin x_1|+|sin x_2|<|x_1|+|x_2|$$



But this doesn't help me much finding a delta...



Thanks for any help!



P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).






calculus functions continuity




calculus functions continuity






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share|cite|improve this question













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edited Jan 1 '13 at 19:48









Haskell Curry

15.2k3886




15.2k3886










asked Jan 1 '13 at 19:27









HaroldHarold

2711315




2711315








  • 1




    $begingroup$
    I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
    $endgroup$
    – Michael Hardy
    Jan 1 '13 at 19:35












  • $begingroup$
    This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
    $endgroup$
    – Julien
    Jan 1 '13 at 19:41
















  • 1




    $begingroup$
    I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
    $endgroup$
    – Michael Hardy
    Jan 1 '13 at 19:35












  • $begingroup$
    This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
    $endgroup$
    – Julien
    Jan 1 '13 at 19:41










1




1




$begingroup$
I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:35






$begingroup$
I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:35














$begingroup$
This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
$endgroup$
– Julien
Jan 1 '13 at 19:41






$begingroup$
This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
$endgroup$
– Julien
Jan 1 '13 at 19:41












4 Answers
4






active

oldest

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18












$begingroup$

Let $epsilon>0$ and $x,yin mathbb{R}$. We want
$$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
Because
$$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
it suffices
$$2left|sinfrac{x-y}2right|<epsilon$$
when
$$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
SInce $left|sin xright|le left|xright|$,
$$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$



Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
    $endgroup$
    – Harold
    Jan 1 '13 at 19:32








  • 1




    $begingroup$
    because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
    $endgroup$
    – Amihai Zivan
    Jan 1 '13 at 19:34








  • 2




    $begingroup$
    Because $left|cos(bullet)right|le 1$.
    $endgroup$
    – Michael Hardy
    Jan 1 '13 at 19:34








  • 1




    $begingroup$
    @Harold $sin x$ is odd
    $endgroup$
    – Nameless
    Jan 1 '13 at 19:48






  • 2




    $begingroup$
    @Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
    $endgroup$
    – Nameless
    Jan 1 '13 at 20:10





















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$begingroup$

By Mean Value Theorem,



$$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$



Hence, you may choose $epsilon=delta$.






share|cite|improve this answer









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  • 4




    $begingroup$
    This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
    $endgroup$
    – wdacda
    Feb 1 '16 at 19:55








  • 2




    $begingroup$
    @wdacda How is that circular reasoning?
    $endgroup$
    – 3x89g2
    Sep 5 '16 at 5:27



















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Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.






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$endgroup$





















    3












    $begingroup$

    There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
    $$|sin(x) - sin(y) | le |x - y|.$$



    This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Another way to prove this identity: math.stackexchange.com/questions/620305/…
      $endgroup$
      – GinKin
      Dec 28 '13 at 10:08










    • $begingroup$
      This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
      $endgroup$
      – ncmathsadist
      Dec 29 '13 at 19:56










    • $begingroup$
      Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
      $endgroup$
      – GinKin
      Dec 29 '13 at 20:03










    • $begingroup$
      What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
      $endgroup$
      – ncmathsadist
      Dec 30 '13 at 0:46













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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    18












    $begingroup$

    Let $epsilon>0$ and $x,yin mathbb{R}$. We want
    $$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
    Because
    $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
    it suffices
    $$2left|sinfrac{x-y}2right|<epsilon$$
    when
    $$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
    SInce $left|sin xright|le left|xright|$,
    $$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$



    Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
      $endgroup$
      – Harold
      Jan 1 '13 at 19:32








    • 1




      $begingroup$
      because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
      $endgroup$
      – Amihai Zivan
      Jan 1 '13 at 19:34








    • 2




      $begingroup$
      Because $left|cos(bullet)right|le 1$.
      $endgroup$
      – Michael Hardy
      Jan 1 '13 at 19:34








    • 1




      $begingroup$
      @Harold $sin x$ is odd
      $endgroup$
      – Nameless
      Jan 1 '13 at 19:48






    • 2




      $begingroup$
      @Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
      $endgroup$
      – Nameless
      Jan 1 '13 at 20:10


















    18












    $begingroup$

    Let $epsilon>0$ and $x,yin mathbb{R}$. We want
    $$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
    Because
    $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
    it suffices
    $$2left|sinfrac{x-y}2right|<epsilon$$
    when
    $$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
    SInce $left|sin xright|le left|xright|$,
    $$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$



    Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
      $endgroup$
      – Harold
      Jan 1 '13 at 19:32








    • 1




      $begingroup$
      because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
      $endgroup$
      – Amihai Zivan
      Jan 1 '13 at 19:34








    • 2




      $begingroup$
      Because $left|cos(bullet)right|le 1$.
      $endgroup$
      – Michael Hardy
      Jan 1 '13 at 19:34








    • 1




      $begingroup$
      @Harold $sin x$ is odd
      $endgroup$
      – Nameless
      Jan 1 '13 at 19:48






    • 2




      $begingroup$
      @Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
      $endgroup$
      – Nameless
      Jan 1 '13 at 20:10
















    18












    18








    18





    $begingroup$

    Let $epsilon>0$ and $x,yin mathbb{R}$. We want
    $$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
    Because
    $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
    it suffices
    $$2left|sinfrac{x-y}2right|<epsilon$$
    when
    $$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
    SInce $left|sin xright|le left|xright|$,
    $$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$



    Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform






    share|cite|improve this answer









    $endgroup$



    Let $epsilon>0$ and $x,yin mathbb{R}$. We want
    $$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
    Because
    $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
    it suffices
    $$2left|sinfrac{x-y}2right|<epsilon$$
    when
    $$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
    SInce $left|sin xright|le left|xright|$,
    $$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$



    Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 1 '13 at 19:29









    NamelessNameless

    10.4k12155




    10.4k12155












    • $begingroup$
      Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
      $endgroup$
      – Harold
      Jan 1 '13 at 19:32








    • 1




      $begingroup$
      because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
      $endgroup$
      – Amihai Zivan
      Jan 1 '13 at 19:34








    • 2




      $begingroup$
      Because $left|cos(bullet)right|le 1$.
      $endgroup$
      – Michael Hardy
      Jan 1 '13 at 19:34








    • 1




      $begingroup$
      @Harold $sin x$ is odd
      $endgroup$
      – Nameless
      Jan 1 '13 at 19:48






    • 2




      $begingroup$
      @Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
      $endgroup$
      – Nameless
      Jan 1 '13 at 20:10




















    • $begingroup$
      Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
      $endgroup$
      – Harold
      Jan 1 '13 at 19:32








    • 1




      $begingroup$
      because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
      $endgroup$
      – Amihai Zivan
      Jan 1 '13 at 19:34








    • 2




      $begingroup$
      Because $left|cos(bullet)right|le 1$.
      $endgroup$
      – Michael Hardy
      Jan 1 '13 at 19:34








    • 1




      $begingroup$
      @Harold $sin x$ is odd
      $endgroup$
      – Nameless
      Jan 1 '13 at 19:48






    • 2




      $begingroup$
      @Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
      $endgroup$
      – Nameless
      Jan 1 '13 at 20:10


















    $begingroup$
    Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
    $endgroup$
    – Harold
    Jan 1 '13 at 19:32






    $begingroup$
    Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
    $endgroup$
    – Harold
    Jan 1 '13 at 19:32






    1




    1




    $begingroup$
    because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
    $endgroup$
    – Amihai Zivan
    Jan 1 '13 at 19:34






    $begingroup$
    because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
    $endgroup$
    – Amihai Zivan
    Jan 1 '13 at 19:34






    2




    2




    $begingroup$
    Because $left|cos(bullet)right|le 1$.
    $endgroup$
    – Michael Hardy
    Jan 1 '13 at 19:34






    $begingroup$
    Because $left|cos(bullet)right|le 1$.
    $endgroup$
    – Michael Hardy
    Jan 1 '13 at 19:34






    1




    1




    $begingroup$
    @Harold $sin x$ is odd
    $endgroup$
    – Nameless
    Jan 1 '13 at 19:48




    $begingroup$
    @Harold $sin x$ is odd
    $endgroup$
    – Nameless
    Jan 1 '13 at 19:48




    2




    2




    $begingroup$
    @Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
    $endgroup$
    – Nameless
    Jan 1 '13 at 20:10






    $begingroup$
    @Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
    $endgroup$
    – Nameless
    Jan 1 '13 at 20:10













    18












    $begingroup$

    By Mean Value Theorem,



    $$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$



    Hence, you may choose $epsilon=delta$.






    share|cite|improve this answer









    $endgroup$









    • 4




      $begingroup$
      This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
      $endgroup$
      – wdacda
      Feb 1 '16 at 19:55








    • 2




      $begingroup$
      @wdacda How is that circular reasoning?
      $endgroup$
      – 3x89g2
      Sep 5 '16 at 5:27
















    18












    $begingroup$

    By Mean Value Theorem,



    $$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$



    Hence, you may choose $epsilon=delta$.






    share|cite|improve this answer









    $endgroup$









    • 4




      $begingroup$
      This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
      $endgroup$
      – wdacda
      Feb 1 '16 at 19:55








    • 2




      $begingroup$
      @wdacda How is that circular reasoning?
      $endgroup$
      – 3x89g2
      Sep 5 '16 at 5:27














    18












    18








    18





    $begingroup$

    By Mean Value Theorem,



    $$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$



    Hence, you may choose $epsilon=delta$.






    share|cite|improve this answer









    $endgroup$



    By Mean Value Theorem,



    $$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$



    Hence, you may choose $epsilon=delta$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 1 '13 at 19:35









    BerkheimerBerkheimer

    1,4371024




    1,4371024








    • 4




      $begingroup$
      This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
      $endgroup$
      – wdacda
      Feb 1 '16 at 19:55








    • 2




      $begingroup$
      @wdacda How is that circular reasoning?
      $endgroup$
      – 3x89g2
      Sep 5 '16 at 5:27














    • 4




      $begingroup$
      This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
      $endgroup$
      – wdacda
      Feb 1 '16 at 19:55








    • 2




      $begingroup$
      @wdacda How is that circular reasoning?
      $endgroup$
      – 3x89g2
      Sep 5 '16 at 5:27








    4




    4




    $begingroup$
    This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
    $endgroup$
    – wdacda
    Feb 1 '16 at 19:55






    $begingroup$
    This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
    $endgroup$
    – wdacda
    Feb 1 '16 at 19:55






    2




    2




    $begingroup$
    @wdacda How is that circular reasoning?
    $endgroup$
    – 3x89g2
    Sep 5 '16 at 5:27




    $begingroup$
    @wdacda How is that circular reasoning?
    $endgroup$
    – 3x89g2
    Sep 5 '16 at 5:27











    12












    $begingroup$

    Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.






    share|cite|improve this answer









    $endgroup$


















      12












      $begingroup$

      Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.






      share|cite|improve this answer









      $endgroup$
















        12












        12








        12





        $begingroup$

        Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.






        share|cite|improve this answer









        $endgroup$



        Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 '13 at 19:59









        Makoto KatoMakoto Kato

        23k560162




        23k560162























            3












            $begingroup$

            There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
            $$|sin(x) - sin(y) | le |x - y|.$$



            This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Another way to prove this identity: math.stackexchange.com/questions/620305/…
              $endgroup$
              – GinKin
              Dec 28 '13 at 10:08










            • $begingroup$
              This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
              $endgroup$
              – ncmathsadist
              Dec 29 '13 at 19:56










            • $begingroup$
              Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
              $endgroup$
              – GinKin
              Dec 29 '13 at 20:03










            • $begingroup$
              What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
              $endgroup$
              – ncmathsadist
              Dec 30 '13 at 0:46


















            3












            $begingroup$

            There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
            $$|sin(x) - sin(y) | le |x - y|.$$



            This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Another way to prove this identity: math.stackexchange.com/questions/620305/…
              $endgroup$
              – GinKin
              Dec 28 '13 at 10:08










            • $begingroup$
              This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
              $endgroup$
              – ncmathsadist
              Dec 29 '13 at 19:56










            • $begingroup$
              Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
              $endgroup$
              – GinKin
              Dec 29 '13 at 20:03










            • $begingroup$
              What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
              $endgroup$
              – ncmathsadist
              Dec 30 '13 at 0:46
















            3












            3








            3





            $begingroup$

            There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
            $$|sin(x) - sin(y) | le |x - y|.$$



            This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.






            share|cite|improve this answer









            $endgroup$



            There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
            $$|sin(x) - sin(y) | le |x - y|.$$



            This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 '13 at 21:03









            ncmathsadistncmathsadist

            43k260103




            43k260103












            • $begingroup$
              Another way to prove this identity: math.stackexchange.com/questions/620305/…
              $endgroup$
              – GinKin
              Dec 28 '13 at 10:08










            • $begingroup$
              This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
              $endgroup$
              – ncmathsadist
              Dec 29 '13 at 19:56










            • $begingroup$
              Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
              $endgroup$
              – GinKin
              Dec 29 '13 at 20:03










            • $begingroup$
              What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
              $endgroup$
              – ncmathsadist
              Dec 30 '13 at 0:46




















            • $begingroup$
              Another way to prove this identity: math.stackexchange.com/questions/620305/…
              $endgroup$
              – GinKin
              Dec 28 '13 at 10:08










            • $begingroup$
              This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
              $endgroup$
              – ncmathsadist
              Dec 29 '13 at 19:56










            • $begingroup$
              Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
              $endgroup$
              – GinKin
              Dec 29 '13 at 20:03










            • $begingroup$
              What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
              $endgroup$
              – ncmathsadist
              Dec 30 '13 at 0:46


















            $begingroup$
            Another way to prove this identity: math.stackexchange.com/questions/620305/…
            $endgroup$
            – GinKin
            Dec 28 '13 at 10:08




            $begingroup$
            Another way to prove this identity: math.stackexchange.com/questions/620305/…
            $endgroup$
            – GinKin
            Dec 28 '13 at 10:08












            $begingroup$
            This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
            $endgroup$
            – ncmathsadist
            Dec 29 '13 at 19:56




            $begingroup$
            This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
            $endgroup$
            – ncmathsadist
            Dec 29 '13 at 19:56












            $begingroup$
            Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
            $endgroup$
            – GinKin
            Dec 29 '13 at 20:03




            $begingroup$
            Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
            $endgroup$
            – GinKin
            Dec 29 '13 at 20:03












            $begingroup$
            What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
            $endgroup$
            – ncmathsadist
            Dec 30 '13 at 0:46






            $begingroup$
            What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
            $endgroup$
            – ncmathsadist
            Dec 30 '13 at 0:46




















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