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Thread: cosine

  1. June 23rd 2008, 12:45 PM #1
    MathNube
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    cosine

    I have no idea where to begin. I'm not used to these kinds of problems.
    Could someone solve it/show work. Please and thank you.

    http://img296.imageshack.us/img296/2...problemmn8.png
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  2. June 23rd 2008, 12:52 PM #2
    icemanfan
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    So you want the value of k such that \cos x = x^2 + k at the value of x = \frac{\pi}{2}. This is the equation: \cos {\frac{\pi}{2}} = \left({\frac{\pi}{2}}\right)^2 + k. This reduces to 0 = \frac{\pi^2}{4} + k. Hence, k = -\frac{\pi^2}{4}.
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  3. June 23rd 2008, 02:13 PM #3
    MathNube
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    does anyone feel like explaining this to me in further detail? i don't really understand it.

    Quote Originally Posted by icemanfan View Post
    So you want the value of k such that \cos x = x^2 + k at the value of x = \frac{\pi}{2}. This is the equation: \cos {\frac{\pi}{2}} = \left({\frac{\pi}{2}}\right)^2 + k. This reduces to 0 = \frac{\pi^2}{4} + k. Hence, k = -\frac{\pi^2}{4}.
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  4. June 23rd 2008, 02:16 PM #4
    topsquark
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    Quote Originally Posted by MathNube View Post
    does anyone feel like explaining this to me in further detail? i don't really understand it.
    For the function to be continuous the values of the two branches of the function must be equal where they meet. Otherwise the function will suffer a "jump" (ie. discontinuity) at x = pi/2. Thus we require
    cos(x) = x^2 + k at x = \frac{\pi}{2}

    And the solution goes on from there as icemanfan wrote.

    -Dan
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  5. June 23rd 2008, 02:16 PM #5
    Isomorphism
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    Quote Originally Posted by MathNube View Post
    does anyone feel like explaining this to me in further detail? i don't really understand it.
    If you clearly specify what is it that you dont understand in icemanfan's explanation, we could help you better.
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  6. June 23rd 2008, 02:34 PM #6
    MathNube
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    ok, why does x=pi/2
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  7. June 23rd 2008, 02:35 PM #7
    topsquark
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    Quote Originally Posted by MathNube View Post
    ok, why does x=pi/2
    Because that is the point where the two branches of the function need to meet. If they are not equal at this point, then the function can't be continuous.

    -Dan
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  8. June 23rd 2008, 04:32 PM #8
    MathNube
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    Are we positive that this is the right answer? According to the person who gave me the problem it is incorrect.
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  9. June 23rd 2008, 05:40 PM #9
    topsquark
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    Quote Originally Posted by MathNube View Post
    Are we positive that this is the right answer? According to the person who gave me the problem it is incorrect.
    Okay. What is the answer you were given and (if you have it) can you post the solution?

    -Dan
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  10. June 23rd 2008, 05:52 PM #10
    MathNube
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    I was not given the solution, just told that it was incorrect.
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  11. June 23rd 2008, 06:05 PM #11
    topsquark
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    Quote Originally Posted by MathNube View Post
    I was not given the solution, just told that it was incorrect.
    I find it hard to argue with the graph. The graphs of the two functions on their respective intervals is clearly continuous with the choice k = -pi^2 / 4. I don't see any way that this solution can be wrong.

    Show this graph to whoever gave you the problem and ask them what their solution looks like. If we are wrong I'd like to see just what the answer is.

    -Dan
    Attached Thumbnails Attached Thumbnails cosine-continuity.jpg  
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