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Math Help - nasty trig induction

  1. #1
    Member jacs's Avatar
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    nasty trig induction

    Here is a nasty trig induction question i dont even pretend to know how to begin. I think it is a little out of the scope of my course, but would like to try it, just in case.


    thanks jacs
    Attached Thumbnails Attached Thumbnails nasty trig induction-inversetaninduction.jpg  
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  2. #2
    Eater of Worlds
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    Add tan^{-1}(\frac{1}{2(n+1)^{2}}) to both sides

    That would be next in your series.

    tan^{-1}(\frac{1}{2\cdot{1^{2}}})+tan^{-1}(\frac{1}{2\cdot{2^{2}}})+ ...........+tan^{-1}(\frac{1}{2n^{2}})+tan^{-1}(\frac{1}{2(n+1)^{2}}) =\frac{\pi}{4}-tan^{-1}(\frac{1}{2n+1})+tan^{-1}(\frac{1}{2(n+1)^{2}})

    Now, your mission is to show that \frac{\pi}{4}-tan^{-1}(\frac{1}{2n+1})+tan^{-1}(\frac{1}{2(n+1)^{2}}) =\frac{\pi}{4}-tan^{-1}(\frac{1}{2(n+1)+1})=\frac{\pi}{4}-tan^{1}(\frac{1}{2n+3})

    It does, but can you show how?.
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  3. #3
    Member jacs's Avatar
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    I am afraid i have no idea how to proceed at all. This question i think is a little beyond what were are supposed to be able to do at our course level. It was presented as a challenge question, but even with your start, i don't know how to proceed.

    jacs
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  4. #4
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    For n=1
    The left hand side is,
    \tan^{-1} \frac{1}{2\cdot 1^2}=\tan^{-1} \frac{1}{2}=\frac{\pi}{4}
    The right hand side is,
    \frac{\pi}{4}-\tan^{-1} \frac{1}{2(1)+1}=\frac{\pi}{4}-\mbox{ some non-zero number}
    No wonder you cannot prove it!
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  5. #5
    Eater of Worlds
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    Quote Originally Posted by ThePerfectHacker
    For n=1
    The left hand side is,
    \tan^{-1} \frac{1}{2\cdot 1^2}=\tan^{-1} \frac{1}{2}=\frac{\pi}{4}
    The right hand side is,
    \frac{\pi}{4}-\tan^{-1} \frac{1}{2(1)+1}=\frac{\pi}{4}-\mbox{ some non-zero number}
    No wonder you cannot prove it!
    Hey PH, I believe you made a slight error.

    tan^{-1}(1)=\frac{\pi}{4}

    Left side:

    tan^{-1}(\frac{1}{2})\neq\frac{\pi}{4}

    It equals 0.463647609001

    The right side:

    \frac{\pi}{4}-tan^{-1}(\frac{1}{3})=0.46347609001

    It is true for n=1
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  6. #6
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    Quote Originally Posted by galactus
    Hey PH, I believe you made a slight error...
    I agree. But I bet you cannot prove me wrong

    Remember mathematicians do not accept numerical, graphical, logical, obvious results
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    I agree. But I bet you cannot prove me wrong

    Remember mathematicians do not accept numerical, graphical, logical, obvious results
    He has proven you wrong - he just has not proven that:

    <br />
\arctan(1/2)=\frac{\pi}{4}-\arctan(1/3)<br />

    which follows immediately from taking tan of both sides and the trig identity
    for the tan of a difference of two angles.

    RonL
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  8. #8
    Eater of Worlds
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    Yes, Cap'N. That method proves the general result, also.

    tan(u-v)=\frac{tan(u)-tan(v)}{1+tan(u)tan(v)}


    Skipping ahead a lot of algebra:

    \frac{\frac{1}{2(n+1)^{2}}-\frac{1}{2n+1}}{1+\frac{1}{2(n+1)^{2}}\cdot\frac{1  }{2n+1}}=\frac{-1}{2n+3}

    tan(\frac{-1}{2n+3})=-tan(\frac{1}{2n+3})
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