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$A= \begin{pmatrix} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x) \\ \end{pmatrix}$

I found that

$A^2 = \begin{pmatrix} \cos^2(x)-\sin^2(x) & 2\cos(x)\sin(x) \\ -2\cos(x)\sin(x) & \cos^2(x)-\sin^2(x) \\ \end{pmatrix}$

$A^2 = \begin{pmatrix} \cos(2x) & \sin(2x) \\ -\sin(2x) & \cos(2x) \\ \end{pmatrix}$

and

$A^3 = \begin{pmatrix} \cos^2(x)[\cos^2(x)-3\sin^2(x)] & \sin(x)[3\cos^2(x)\sin^2(x)] \\ -\sin(x)[3\cos^2(x)+\sin^2(x)] & \cos(x)[-3\sin^2(x)+\cos^2(x)] \\ \end{pmatrix}$ $A^3 = \begin{pmatrix} \cos(2x)cos(x) - \sin(x)\sin(2x) & \sin(x)\cos(2x) + \cos(x)\sin(2x) \\ -\cos(x)sin(2x)-\sin(x)\cos(2x) & \cos(x) + \cos(2x) -\sin(x)\sin(2x) \\ \end{pmatrix}$

but I am unable to find a general solution. I probably need to perform some algebraic manipulations or whatnot.

wonton
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  • Try using trig identities; e.g. For the (1,1) entry: $\cos(2x)\cos(x)-\sin(2x)\sin(x)=\cos(3x)$. – Daryl Feb 05 '13 at 06:42

2 Answers2

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Hint: this is rotation by the angle x

Alex Youcis
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In general, following formula holds: (Proof of this formula uses induction.) $$A^n =\begin{pmatrix} \cos (nx) & -\sin(nx) \\ \sin(nx) & \cos(nx) \end{pmatrix}$$

In fact, matrix $\begin{pmatrix} \cos (x) & -\sin(x) \\ \sin(x) & \cos(x) \end{pmatrix}$ behave similar to $\cos x + i \sin x$.

Hanul Jeon
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  • do you know how to transform A^3 to the 3x form? – wonton Feb 05 '13 at 06:16
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    @wonton You can use the angle addition identities for sine and cosine applied to the last formula in your question. The more fundamental reason is the one Alex Yocuis mentioned however; the matrix behaves this way because applying a rotation of angle $x$ a total of $n$ times is the same as applying a single rotation of $nx$. – EuYu Feb 05 '13 at 06:26