Facebook Page
Twitter
RSS
+ Reply to Thread
Page 1 of 2 12 LastLast
Results 1 to 10 of 12
  1. MHB Craftsman

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    350
    Thanks
    106 times
    Thanked
    59 times
    #1
    Hello all

    I am trying to prove that

    \[sin(x)-cos(x)\geq 1\]

    For each x in the interval \[[\frac{\pi }{2},\pi ]\]

    I tried doing it by contradiction, what I did was:

    Assume

    \[sin(x)-cos(x)< 1\]

    Then I used the power of 2 on each side of the inequality and got:

    \[sin^{2}(x)-2sin(x)cos(x)+cos^{2}(x)<1\]


    which led me to

    \[0<2sin(x)cos(x)\]

    which is a contradiction since

    \[cos(\frac{\pi }{2})=0\]

    I am not sure that what I did is correct or complete. Can you please check my proof and give me your opinion on the matter?

    Thank you in advance.

  2. MHB Apprentice

    Status
    Offline
    Join Date
    Jan 2017
    Posts
    17
    Thanks
    49 times
    Thanked
    8 times
    #2
    What you did upto the last but one step is perfect, but your showing the contradiction is not that correct, as $\cos (\frac{\pi}{2})=0$ is true only for one value, i.e. at $\pi/2$. The correct contradiction comes from the fact that $2\sin x \cos x\le0 \ \forall x\in[\frac{\pi}{2},\pi]$ from the non-positive nature of $\cos $ function.

  3. MHB Craftsman

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    350
    Thanks
    106 times
    Thanked
    59 times
    #3 Thread Author
    Thank you. One question if I may.

    Logically, if I need to show that for all point in the interval this statement is true, and I showed that there exists a point for which it is not, isn't it enough?

  4. Perseverance
    MHB Global Moderator
    MHB Math Helper
    greg1313's Avatar
    Status
    Online
    Join Date
    Feb 2013
    Location
    London, Ontario, Canada - The Forest City
    Posts
    951
    Thanks
    3,662 times
    Thanked
    1,906 time
    Thank/Post
    2.004
    #4
    Quote Originally Posted by Yankel View Post
    Logically, if I need to show that for all point in the interval this statement is true, and I showed that there exists a point for which it is not, isn't it enough?
    Your conclusion only shows that the original statement must be true at $x=\frac{\pi}{2}$. What about the remainder of the interval? Does your conclusion tell us anything about that?

  5. MHB Craftsman

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    350
    Thanks
    106 times
    Thanked
    59 times
    #5 Thread Author
    I see what you mean ! You are right !

    I know notice that my proof is wrong !

    \[a<b \sim \rightarrow a^{2}<b^{2}\]

    My first step is wrong, isn't it?
    Last edited by Yankel; January 14th, 2017 at 09:00.

  6. MHB Craftsman

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    350
    Thanks
    106 times
    Thanked
    59 times
    #6 Thread Author
    by the way, is there a way of proving it without using contradiction?

    If I know that sin(x) in this interval is between 1 to 0, and cos(x) between 0 to -1, can I claim that this is enough to be considered a proof?

  7. Perseverance
    MHB Global Moderator
    MHB Math Helper
    greg1313's Avatar
    Status
    Online
    Join Date
    Feb 2013
    Location
    London, Ontario, Canada - The Forest City
    Posts
    951
    Thanks
    3,662 times
    Thanked
    1,906 time
    Thank/Post
    2.004
    #7
    Your proof is ok, it's your conclusion that needs work. See vidyarth's post (post #2).

    As an alternative, write $\sin(x)-\cos(x)$ as $\sqrt2\sin\left(x-\frac{\pi}{4}\right)$ and note the properties of this sine function over the given interval.

  8. MHB Journeyman
    MHB Site Helper
    MHB Math Helper
    Rido12's Avatar
    Status
    Online
    Join Date
    Jul 2013
    Posts
    698
    Thanks
    3,330 times
    Thanked
    806 times
    Thank/Post
    1.155
    Awards
    Chat Box Champion (2014)  

MHB Model User Award (2014)
    #8
    Quote Originally Posted by Yankel View Post
    I see what you mean ! You are right !

    I know notice that my proof is wrong !

    \[a<b \sim \rightarrow a^{2}<b^{2}\]

    My first step is wrong, isn't it?
    Yes, $a<b$ does not imply $a^2<b^2$ in generality. Consider $a=-4$ and $b=2$. I haven't looked at the problem, but if you want to use that line of thinking, you will need to consider the cases when (a) $a, b>0$, (b) $a, b <0$, (c), ...etc. In particular, for the case of (b), we have $a<b \implies a^2>b^2$.
    Last edited by Rido12; January 15th, 2017 at 03:47.

  9. Perseverance
    MHB Global Moderator
    MHB Math Helper
    greg1313's Avatar
    Status
    Online
    Join Date
    Feb 2013
    Location
    London, Ontario, Canada - The Forest City
    Posts
    951
    Thanks
    3,662 times
    Thanked
    1,906 time
    Thank/Post
    2.004
    #9
    Quote Originally Posted by Rido12 View Post
    Yes, $a<b$ does not imply $a^2<b^2$ in generality. Consider $a=-4$ and $b=2$. I haven't looked at the problem, but if you want to use that line of thinking, you will need to consider the cases when (a) $a, b>0$, (b) $a, b <0$, (c), ...etc. In particular, for the case of (b), we have $a<b \implies a^2>b^2$.
    Yes; quite right. My apologies for my blunder.

  10. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    I like Serena's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,261
    Thanks
    4,207 times
    Thanked
    11,742 times
    Thank/Post
    1.875
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #10
    We can use the triangle inequality.
    \begin{tikzpicture}
    [blue, ultra thick]
    \draw (0,0) -- node[above right] {1} (-3,4) -- node[left] {$\sin x$} (-3, 0) -- node[below] {$-\cos x$} cycle;

    \end{tikzpicture}

    The triangle inequality says that the sum of two sides of a triangle is greater than or equal to the third side.
    So:
    $$\sin x + (-\cos x) \ge 1$$

Similar Threads

  1. Replies: 9
    Last Post: December 30th, 2016, 22:19
  2. Prove the sum of a and d is equal the sum of b and c.
    By anemone in forum Challenge Questions and Puzzles
    Replies: 5
    Last Post: January 19th, 2016, 23:57
  3. Prove the product is less than or equal to 1
    By anemone in forum Challenge Questions and Puzzles
    Replies: 2
    Last Post: November 8th, 2015, 01:50
  4. Prove y is less than or equal to z
    By anemone in forum Challenge Questions and Puzzles
    Replies: 2
    Last Post: May 28th, 2015, 23:17
  5. Prove that the sum is equal to -1.
    By evinda in forum Number Theory
    Replies: 7
    Last Post: October 6th, 2014, 23:42

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards