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Math Help - Mathematical Induction HELP!

  1. #1
    Zel
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    Question Mathematical Induction HELP!

    Use the principle of mathematical induction to prove the following:
    cosθ + cos3θ + cos5θ ... + cos(2n-1)θ = [sin 2nθ]/[2sinθ] for all positive n

    sinθ + sin3θ + sin5θ ... + sin(2n-1)θ = [1-cos2nθ]/[2sinθ] for all positive n

    1 + cisθ + cis2θ ... + cis nθ = [1-cis(n+1)θ] / [1-cisθ] where cisθ doesn't equal 1, n is greater than or equal to 1

    I have solved these for n=1, so i said P(k)=1 is true (n=k), but now I am onto P(k)= k+1, and I am really stuck!
    Any help would be greatly appreciated!
    Thanks in advance =]
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  2. #2
    MHF Contributor
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    Hello Zel
    Quote Originally Posted by Zel View Post
    Use the principle of mathematical induction to prove the following:
    cosθ + cos3θ + cos5θ ... + cos(2n-1)θ = [sin 2nθ]/[2sinθ] for all positive n

    sinθ + sin3θ + sin5θ ... + sin(2n-1)θ = [1-cos2nθ]/[2sinθ] for all positive n

    1 + cisθ + cis2θ ... + cis nθ = [1-cis(n+1)θ] / [1-cisθ] where cisθ doesn't equal 1, n is greater than or equal to 1

    I have solved these for n=1, so i said P(k)=1 is true (n=k), but now I am onto P(k)= k+1, and I am really stuck!
    Any help would be greatly appreciated!
    Thanks in advance =]
    You sound as though you have the correct structure for the induction proofs, so here are the trig bits that you need to show that P(k) \rightarrow P(k+1).

    1) You'll need to show that:
    \frac{\sin 2k\theta}{2\sin\theta} + \cos(2k+1)\theta = \frac{\sin2(k+1)\theta}{2\sin\theta}
    So:
    \frac{\sin 2k\theta}{2\sin\theta} + \cos(2k+1)\theta =  \frac{\sin2k\theta + 2\sin\theta\cos(2k+1)\theta}{2\sin\theta}
    =   \frac{\sin2k\theta + 2\sin\theta(\cos2k\theta\cos\theta - \sin2k\theta\sin\theta)}{2\sin\theta}

    =   \frac{\sin2k\theta(1-2\sin^2\theta) + 2\sin\theta\cos2k\theta\cos\theta}{2\sin\theta}


    =   \frac{\sin2k\theta\cos2\theta + \cos2k\theta\sin2\theta}{2\sin\theta}


    = \frac{\sin2(k+1)\theta}{2\sin\theta}

    2) is very similar. Start in the same way, combining to a single fraction. Then expand \sin(2k+1)\theta and group the resulting terms in \cos2k\theta.

    3) is dead easy if you remember that:
    \cos n\theta+i\sin n\theta =(\cos \theta + i \sin \theta)^n
    and hence:
    \cos (n+1)\theta+i\sin (n+1)\theta =(\cos \theta + i \sin \theta)^{n+1}
    =(\cos \theta + i \sin \theta)^n(\cos\theta + i\sin\theta)
    And you will need to use it in the form:
    \text{cis} (n+2)\theta = \text{cis} (n+1) \theta \text{ cis}\theta
    Can you complete it now?

    Grandad
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