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Math Help - Proof by induction

  1. #1
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    Proof by induction

    This problem recently came up in an FP3 (old P6) textbook, and none of my 7-strong further maths class have been able to solve it

    Prove by induction that:

    <br />
\sum_{r=1}^n\ \tan(r\theta)\tan((r+1)\theta)=\tan((n+1)\theta) \cot(\theta)-n-1<br />

    Thanks for any help you can give; this problem has really started to irritate us!
    Last edited by CaptainBlack; April 29th 2006 at 06:09 AM. Reason: To make the problem more readable
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  2. #2
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    Quote Originally Posted by rabywebb
    This problem recently came up in an FP3 (old P6) textbook, and none of my 7-strong further maths class have been able to solve it

    Prove by induction that Σ from r = 1 to n of tan rΘ tan (r + 1)Θ = tan (n + 1)Θ cotΘ - n - 1

    Thanks for any help you can give; this problem has really started to irritate us!
    This will only be a sketch of the proof, I will leave the demonstration of the
    truth of the base case to the reader, as well as the weasle words needed
    to make this a proof by induction.

    We are asked to prove:

    <br />
\sum_{r=1}^n\ \tan(r\theta)\tan((r+1)\theta)=\tan((n+1)\theta) \cot(\theta)-n-1<br />

    Let (for convienience only):

    <br />
L_n(\theta)=\sum_{r=1}^n\ \tan(r\theta)\tan((r+1)\theta)<br />

    and

    <br />
R_n(\theta)=\tan((n+1)\theta) \cot(\theta)-n-1<br />

    So we now need to prove that:

    <br />
L_n(\theta)=R_n(\theta)<br />

    Suppose this is true for n=k then we need to prove that it is true
    for n=k+1.

    Now:

    <br />
L_{k+1}(\theta)=L_k(\theta) +\tan((k+1)\theta)\tan((k+2)\theta) =R_k(\theta) +\tan((k+1)\theta)\tan((k+2)\theta)<br />

    and:

    <br />
R_{k+1}(\theta)=\tan((k+2)\theta) \cot(\theta)-k-2<br />

    Thus to prove what is to be proven for k+1 is equivalent to proving that:

    <br />
\tan((k+2)\theta) \cot(\theta)-k-2= R_k(\theta) +\tan((k+1)\theta)\tan((k+2)\theta)<br />
,

    or:

    <br />
\tan((k+2)\theta) \cot(\theta)-k-2= \tan((k+1)\theta) \cot(\theta)-k-1 +\tan((k+1)\theta)\tan((k+2)\theta)<br />
.

    Which may be simplified to:

    <br />
\tan((k+2)\theta) \cot(\theta)-1= \tan((k+1)\theta) \cot(\theta) +\tan((k+1)\theta)\tan((k+2)\theta)\ \ \ \ \dots\ (1)<br />
.

    To prove this (there must be many other ways) use the trig-identity for
    the \tan of a sum to obtain:

    <br />
\tan((k+2) \theta)=\frac{\tan((k+1)\theta)+\tan(\theta)}{1-\tan((k+1)\theta)\tan(\theta)}<br />
\ \ \ \ \ \dots\ (2)<br />
.

    Now substitute the RHS of (2) for  \tan((k+2)\theta) on both the RHS and LHS of (1),
    on simplification these are both:

    <br />
\frac{\tan((k+1)\theta)[\cot(\theta)+\tan(\theta)]}{1+\tan((k+1)\theta)\tan(\theta)}<br />
.

    Which proves:

    <br />
L_{k+1}(\theta)=R_{k+1}(\theta)<br />

    from the assumption:

    <br />
L_k(\theta)=R_k(\theta)<br />
.

    RonL
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  3. #3
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    Thank you very much!
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by rabywebb
    Thank you very much!
    Your welcome

    RonL
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