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Math Help - Proving by induction

  1. #1
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    Proving by induction

    If sinx cannot = 0, use mathematical induction to show that

    cosx \cdot cos2x \cdot cos4x... cos2^{n-1} x = \frac{sin2^n x}{2^n sinx} for every integer n \geq 1

    So for the induction step what I have so far is:

    Assume n=k is true, show true for n=k+1

    cosx \cdot cos2x \cdot cos4x... cos2^{k-1} x \cdot cos2^{(k+1)-1}x = \frac{sin2^{k+1} x}{2^{k+1} sinx}

    \frac{sin2^{k}x}{2^{k}sinx} \cdot cos2^{k}x = \frac{sin2^{k+1}x}{2^{k+1}sinx}

    \frac{(sin2^{k}x)(cos2^{k}x)}{2^{k}sinx} = \frac{sin2^{k+1}x}{2^{k+1}sinx}

    Assuming any of this is correct so far, I am having trouble showing that both sides are equal from here on.
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  2. #2
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    Re: Proving by induction

    Note that sin(2x) = 2sin(x)cos(x).
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  3. #3
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    Re: Proving by induction

    So then I'll have:

    \frac{(sin2^{k}x)(cos2^{k}x)}{2^{k}sinx} = \frac{2sin^{k+1}x cos^{k+1}x}{2^{k+1}sinx}

    Is that what you meant I should do next? Where do I go from here then?
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    Re: Proving by induction

    Quote Originally Posted by BobRoss View Post
    So then I'll have:

    \frac{(sin2^{k}x)(cos2^{k}x)}{2^{k}sinx} = \frac{2sin^{k+1}x cos^{k+1}x}{2^{k+1}sinx}

    Is that what you meant I should do next?
    No, the right-hand side is wrong.. We have \sin(2^{k+1}x)=\sin(2\cdot2^kx)=2\sin(2^kx)\cos(2^  kx).
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  5. #5
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    Re: Proving by induction

    Quote Originally Posted by emakarov View Post
    No, the right-hand side is wrong.. We have \sin(2^{k+1}x)=\sin(2\cdot2^kx)=2\sin(2^kx)\cos(2^  kx).
    Why does \sin(2^{k+1}x)=\sin(2\cdot2^kx)

    ?
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  6. #6
    Junior Member DIOGYK's Avatar
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    Re: Proving by induction

    Quote Originally Posted by BobRoss View Post
    Why does \sin(2^{k+1}x)=\sin(2\cdot2^kx)

    ?
    For example:
    a^2*a^3=a^{2+3}=a^{5}..
    Similarly,
    \sin(2^1*2^kx)=\sin(2^{k+1}x)
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  7. #7
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    Re: Proving by induction

    Oh okay that makes sense. Then I can do the same thing with the term 2^{k+1}sinx to get  2 \cdot 2^{k}sinx And doing that makes the left side equal the right side, and that proves the equation, correct?
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  8. #8
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    Re: Proving by induction

    Quote Originally Posted by BobRoss View Post
    Oh okay that makes sense. Then I can do the same thing with the term 2^{k+1}sinx to get  2 \cdot 2^{k}sinx And doing that makes the left side equal the right side, and that proves the equation, correct?
    Yes.
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