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Math Help - Trigonometric Matrixes?

  1. #1
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    Unhappy Trigonometric Matrixes?

    I need to prove by induction that this 2x2 matrix:

    [cosθ -senθ; senθ cosθ]^n is equal to this 2x2 matrix:

    [cos(n)θ -sen(n)θ; sen(n)θ cos(n)θ] for all n ∈ Z.

    Since it's by induction, I've already got the n = 1 and n = k, but... now what?

    Thanks!
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  2. #2
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    Quote Originally Posted by El Joey View Post
    I need to prove by induction that this 2x2 matrix:

    [cosθ -senθ; senθ cosθ]^n is equal to this 2x2 matrix:

    [cos(n)θ -sen(n)θ; sen(n)θ cos(n)θ] for all n ∈ Z.

    Since it's by induction, I've already got the n = 1 and n = k, but... now what?

    Thanks!
    Here's the idea:

    Let Q(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin \theta & \cos\theta \end{pmatrix}

    Show that Q(\theta)Q(\alpha) = Q(\theta + \alpha), by multiplying out.

    Now choose \alpha = n\theta and use the induction hypothesis to finish the proof
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by El Joey View Post
    I need to prove by induction that this 2x2 matrix:

    [cosθ -senθ; senθ cosθ]^n is equal to this 2x2 matrix:

    [cos(n)θ -sen(n)θ; sen(n)θ cos(n)θ] for all n ∈ Z.

    Since it's by induction, I've already got the n = 1 and n = k, but... now what?

    Thanks!
    Let A=\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

    Assume that A^k=\begin{pmatrix} \cos k \theta & - \sin k \theta \\ \sin k \theta & \cos k \theta \end{pmatrix}

    Then A^{k+1}=A^kA=\begin{pmatrix} \cos k \theta & - \sin k \theta \\ \sin k \theta & \cos k \theta \end{pmatrix} \begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

    Now make the multiplication and use these formula :
    \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)
    \sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)

    and you're done
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