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Yet another induction proof

  1. Feb 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove the following using induction:
    [tex]cos(n\pi + x) = (-1)^ncos(x) [/tex]

    3. The attempt at a solution
    Let [tex] cos(k\pi + x) = (-1)^k cos(x)[/tex]
    True for k=1

    Right Hand Side...
    [tex]= -(-1)^k cos(x)[/tex]
    [tex]= (-1)^k cos(x)[/tex]

    Since [tex](-1)^{k+1}cos(x) = (-1)^k cos(x)[/tex]...
    Not too sure where to go from here.
    Last edited: Feb 11, 2007
  2. jcsd
  3. Feb 11, 2007 #2


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    Why not take the LHS [itex]cos((k+1)\pi +x)=cos([k\pi +x]+\pi)[/itex] and use the double angle formula: cos(a+b)=cosa cosb-sina sinb?
  4. Feb 11, 2007 #3

    matt grime

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    Are you claiming that's a proof for k=1, an attempt at a proof for k=1, or are you stating it is true for k=1 without any proof?

    Anyway, this is about proof by induction, and obviously the left hand side is the thing you're going to want to attack inductively (since there is an obvious way to write k+1 as k plus 1 and you know you're trig formulae (cos(A+B)=cos(A)cos(B)-sin(A)sin(B) for instance...)
  5. Feb 15, 2007 #4
    I did the LHS, and got

    [itex]cos((k+1)\pi +x) = cos(k\pi +x)[/itex]

    and from before,

    [tex](-1)^{k+1}cos(x) = (-1)^k cos(x)[/tex]

    And so from that, and the initial statement, does that prove it?
  6. Feb 15, 2007 #5


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    err.. isn't [tex](-1)^{k+1}\cos(x) = (-1) (-1)^k \cos(x)[/tex]

    so using your assumption:

    ans so on... all too easy :smile:
  7. Feb 15, 2007 #6

    matt grime

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    this is wrong: look at the graph of cos(x).

    this is also wrong: you've just said 1=-1.
  8. Feb 15, 2007 #7
    Oops, silly mistake.

    So let me try this again,

    [tex]cos(k\pi + x) = (-1)^kcos(x)[/tex]

    [tex](-1)^{k+1}cosx = -(-1)^{k}cosx [/tex]

    [tex]cos[(k+1)\pi + x] = cos[(k\pi + x) + \pi][/tex]
    using double angle formula,
    [tex]= -cos(k\pi + x)[/tex]

    And so, [tex]cos[(k+1)\pi + x] = (-1)^{k+1}cosx[/tex]
    => [tex]-cos(k\pi + x) = -(-1)^{k}cosx[/tex]

    kill the negatives

    [tex]cos(k\pi + x) = (-1)^{k}cosx[/tex]

    Last edited: Feb 15, 2007
  9. Feb 15, 2007 #8


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    Here, you've assumed that the statement holds for n=k+1, and used this to prove for n=k. This to me doesn't seem right, as surely you need to show that the statement holding for n=k => the statement is true for n=k+1.

    To do this, we suppose that [itex]\cos(k\pi+x)=(-1)^k\cos(x)[/itex] is true. Now, we use this to prove that it hold for n=k+1; [tex]\cos([k+1]\pi+x)=\cos([k\pi+x]+\pi)=-\cos(k\pi+x)=-(-1)^k\cos(x)=(-1)^{k+1}\cos(x)[/tex] , where we note that the penultimate equality holds from our inductive hypothesis that [itex]\cos(k\pi+x)=(-1)^k\cos(x)[/itex]
    Last edited: Feb 15, 2007
  10. Feb 15, 2007 #9

    matt grime

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    What are you doing? LHS and RHS of what? Not what you wrote one line above.

    Induction proofs are taught by rote. So reproduce the style of proof.

    And don't write things going from both sides like this. Sure, that's often how you work out what's going on (to get from A to B yo go from A to C and from B to C as well). What you write should be a simple argument with no jumping around:

    It's trivially true for n=1.

    Consider cos(pi(k+1)+x), and simplify it and now use the inductive hypothesis that the result is true for k (and everything between 1 and k as well if necessary - it isn't in this case).
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