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Math Help - Which is larger?

  1. #1
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    Which is larger?

    Let x be an element of R. Find which term has a larger value, sin(cos x) or cos(sin x)?


    Any help or detailed working out would be appreciated!
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    Quote Originally Posted by Mr Rayon View Post
    Let x be an element of R. Find which term has a larger value, sin(cos x) or cos(sin x)?


    Any help or detailed working out would be appreciated!
    Dear Mr Rayon,

    Please refer the attached file. The blue line represents y=sin(cosx) while the green curve represents y=cos(sinx). From this graph it is clear that sin(cosx)<cos(sinx). But for the moment I dont have any idea of how prove this result.
    Attached Thumbnails Attached Thumbnails Which is larger?-sp.png  
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  3. #3
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    Quote Originally Posted by Mr Rayon View Post
    Let x be an element of R. Find which term has a larger value, sin(cos x) or cos(sin x)?


    Any help or detailed working out would be appreciated!
    Here is another view.

    sinx ranges from -1\rightarrow\ 1

    cos(sinx) therefore ranges from cos(-1)=cos(1)=0.54\rightarrow\ cos(0)=1

    cosx ranges from -1\rightarrow\ 1

    sin(cosx) therefore ranges from sin(-1)=-0.84\rightarrow\ sin(1)=0.84

    A little more analysis will show they are never equal.
    Attached Thumbnails Attached Thumbnails Which is larger?-sin-cosx-.jpg  
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    Quote Originally Posted by Archie Meade View Post
    Here is another view.

    sinx ranges from -1\rightarrow\ 1

    cos(sinx) therefore ranges from cos(-1)=cos(1)=0.54\rightarrow\ cos(0)=1

    cosx ranges from -1\rightarrow\ 1

    sin(cosx) therefore ranges from sin(-1)=-0.84\rightarrow\ sin(1)=0.84

    A little more analysis will show they are never equal.
    Dear Archie Meade,

    Just out of curiosity, with what software did you create your images???
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    Dear Archie Meade,

    Just out of curiosity, with what software did you create your images???
    Hi Sudharaka,

    I only use the "pages" application in Apple iWorks!
    in conjunction with a Mathtype program (very basic).

    AM
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  6. #6
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    Quote Originally Posted by Mr Rayon View Post
    Let x be an element of R. Find which term has a larger value, sin(cos x) or cos(sin x)?


    Any help or detailed working out would be appreciated!
    Both functions are continuous and you can't find a value of x for which they are equal.
    So, if one is bigger than the other for a particular value of x, it will be bigger for all values of x.
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  7. #7
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    Here is a way we can develop the proof...

    cos(sinx)>0

    hence we need not be concerned with sin(cosx)<0

    and the maximum value of sin(cosx) is <1.

    We only need consider both functions >0 which is in quadrant 1, where 0<x<\frac{\pi}{2}

    In quadrant 1, \displaystyle\ sin(cosx)=cos\left(\frac{\pi}{2}-cosx\right)

    Hence we are examining \displaystyle\ cos(sinx)-cos\left(\frac{\pi}{2}-cosx\right)

    so we want to know if sinx<\left[\frac{\pi}{2}-cosx\right]

    since in quadrant 1, cos(angle)>cos(larger\;angle)

    \frac{\pi}{2}>sinx+cosx ?

    Hence we find the maximum value of sinx+cosx by differentiating...

    cosx-sinx=0\Rightarrow\ cosx=sinx which happens at x=\frac{\pi}{4} radians.

    The maximum value of sinx+cosx=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\f  rac{2}{\sqrt{2}}=\sqrt{2}

    \frac{\pi}{2}>\sqrt{2}\Rightarrow\ cos(sinx)>sin(cosx)
    Attached Thumbnails Attached Thumbnails Which is larger?-cos-sinx-sin-cosx-.jpg  
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    Quote Originally Posted by BobP View Post
    Both functions are continuous and you can't find a value of x for which they are equal.
    So, if one is bigger than the other for a particular value of x, it will be bigger for all values of x.
    Hi BobP,

    Is this a theorem? Can you please tell me where you have encountered it?
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    Quote Originally Posted by Sudharaka View Post
    Hi BobP,

    Is this a theorem? Can you please tell me where you have encountered it?
    If it isn't a theorem I'm happy to claim it !

    Actually, I thought it was just common sense .
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  10. #10
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    Hi everyone,

    Below is a proof that I have formulated for the theorem BobP had given in post #6. If you find any mistake in it please do not hesitate to tell me. Using this result the above problem could be easily solved.
    Attached Thumbnails Attached Thumbnails Which is larger?-untitled-1.pdf  
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  11. #11
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    The graph of the two functions sin(cosx) and cos(sinx) or cos(sinx)-sin(cosx) are a nice guide.

    If we are presented with just the 2 functions, cos(sinx) and sin(cosx)

    and we want to compare them, how do we answer "what stands out ?"
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  12. #12
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    Quote Originally Posted by Archie Meade View Post
    The graph of the two functions sin(cosx) and cos(sinx) or cos(sinx)-sin(cosx) are a nice guide.

    If we are presented with just the 2 functions, cos(sinx) and sin(cosx)

    and we want to compare them, how do we answer "what stands out ?"
    Dear Archie Meade,

    I dont quite understand what you meant by saying ".......how do we answer what stands out?" Can you please explain this.
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  13. #13
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    Which is larger?

    Quote Originally Posted by BobP View Post
    Both functions are continuous and you can't find a value of x for which they are equal.
    So, if one is bigger than the other for a particular value of x, it will be bigger for all values of x.

    This tread is getting old, but Mr. Ryan still seemed confused.

    This may be a Lemma to BobP's theorem.

    I hope you can agree that: If two functions, f(x) and g(x) are continuous, and if they are not equal for any x, then one of them is greater than the other for all x.

    In this case let's see if we can solve:

    \sin(\cos(x))=\cos(\sin(x)).

    \cos(\theta)=\sin(\theta+{\pi\over2}),

    so \sin(\cos(x))=\sin(\sin(x)+{\pi\over2}).

    Then \cos(x)=\sin(x)+{\pi\over2}. (O.K., you could add integer multiple of 2\pi to this, but that won't help.)

    \cos(x)-\sin(x)={\pi\over2}

    Use the identity: \cos(x)-\sin(x)=\sqrt{2}\cos(x+{\pi\over4}).

    This gives us \sqrt{2}\cos(x+{\pi\over4})={\pi\over2}

    The maximum value of the cosine is 1, and \sqrt{2}<{\pi\over2}, so \sin(\cos(x))=\cos(\sin(x)) has no solution.

    Since \sin(\cos(\pi))=\sin(-1)\approx -0.8415 (It's definitely negative!)

    and \cos(\sin(\pi))=\cos(0)=1,

    it must be that \sin(\cos(x))<\cos(\sin(x)), for all x.


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  14. #14
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    Maybe simplest and without reference to graphs is:

    cos(A+B)=cosAcosB-sinAsinB\Rightarrow\ cos\left(\frac{\pi}{2}-A\right)=sinA

    cos(sinx)-sin(cosx)=cos(sinx)-cos\left(\frac{\pi}{2}-cosx\right)

    This is the difference of cosines.

    cosA-cosB=-2sin\left[\frac{A+B}{2}\right]sin\left[\frac{A-B}{2}\right]

    \Rightarrow\ cos(sinx)-sin(cosx)=-2sin\left[\frac{sinx-cosx+\frac{\pi}{2}}{2}\right]sin\left[\frac{sinx+cosx-\frac{\pi}{2}}{2}\right]

    One could go further with trigonometry here.
    An alternative is to examine the contents of the square brackets on the previous line.
    We can examine the maximum and minimum values of these expressions.

    \frac{d}{dx}(sinx-cosx)=cosx+sinx=0\Rightarrow\ cosx=-sinx

    Referring to the unit-circle, this occurs when x=\frac{3{\pi}}{4},\;\;x=\frac{7{\pi}}{4}

    (1) maximum value within the first square brackets

    x=\frac{3{\pi}}{4}\Rightarrow\frac{\frac{1}{\sqrt{  2}}+\frac{1}{\sqrt{2}}+\frac{\pi}{2}}{2}=\frac{1}{  \sqrt{2}}+\frac{\pi}{4}

    0<\frac{1}{\sqrt{2}}+\frac{\pi}{4}<\frac{\pi}{2}\R  ightarrow\ sin\left[\frac{sinx-cosx+\frac{\pi}{2}}{2}\right]>0

    (2) minimum value within the first square brackets

    x=\frac{7{\pi}}{4}\Rightarrow\frac{-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+\frac{\pi}{2}}{2}=-\frac{1}{\sqrt{2}}+\frac{\pi}{4}

    0<-\frac{1}{\sqrt{2}}+\frac{\pi}{4}<\frac{\pi}{2}\Rig  htarrow\ sin\left[\frac{sinx-cosx+\frac{\pi}{2}}{2}\right]>0

    \frac{d}{dx}(sinx+cosx)=cosx-sinx=0\Rightarrow\ cosx=sinx\Rightarrow\ x=\frac{\pi}{4},\;\;x=\frac{5{\pi}}{4}

    (3) maximum value within the second square brackets

    x=\frac{\pi}{4}\Rightarrow\frac{\frac{1}{\sqrt{2}}  +\frac{1}{\sqrt{2}}-\frac{\pi}{2}}{2}=\frac{1}{\sqrt{2}}-\frac{\pi}{4}

    -\frac{\pi}{2}<\frac{1}{\sqrt{2}}-\frac{\pi}{4}<0\Rightarrow\ sin\left[\frac{sinx+cosx-\frac{\pi}{2}}{2}\right]<0

    (4) minimum value within the second square brackets

    x=\frac{5\pi}{4}\Rightarrow\frac{-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{\pi}{2}}{2}=-\frac{1}{\sqrt{2}}-\frac{\pi}{4}

    -\frac{\pi}{2}<-\frac{1}{\sqrt{2}}-\frac{\pi}{4}<0\Rightarrow\ sin\left[\frac{cosx+sinx-\frac{\pi}{2}}{2}\right]<0

    Therefore, it follows that

    cos(sinx)-sin(cosx)>0\Rightarrow\ cos(sinx)>sin(cosx)
    Last edited by Archie Meade; January 1st 2011 at 03:50 AM. Reason: typo
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