# Thread: Proof of the rotation matrix

1. ## Proof of the rotation matrix

Hello. Somebody knows some proof for the
cosθ-sinθ
sinθ cosθ
rotation matrix? Ι can't compromize for the minus sign.
Thanks.

2. ## Re: Proof of the rotation matrix

Say you want to rotate a point \displaystyle \displaystyle \begin{align*} (x, y) \end{align*} to \displaystyle \displaystyle \begin{align*} (x', y') \end{align*} by an angle of \displaystyle \displaystyle \begin{align*} \alpha \end{align*} in the anticlockwise direction. Notice that if we switch to polars, we have \displaystyle \displaystyle \begin{align*} x = r\cos{(\theta)} \end{align*} and \displaystyle \displaystyle \begin{align*} y = r\sin{(\theta)} \end{align*}. When these are rotated by an angle of \displaystyle \displaystyle \begin{align*} \alpha \end{align*}, that gives \displaystyle \displaystyle \begin{align*} x' = r\cos{(\theta + \alpha)} \end{align*} and \displaystyle \displaystyle \begin{align*} y = r\sin{(\theta + \alpha)} \end{align*}.

Simplifying these using the identities \displaystyle \displaystyle \begin{align*} \sin{(A + B)} \equiv \sin{(A)}\cos{(B)} + \cos{(A)}\sin{(B)} \end{align*} and \displaystyle \displaystyle \begin{align*} \cos{(A + B)} \equiv \cos{(A)}\cos{(B)} - \sin{(A)}\sin{(B)} \end{align*} we get

\displaystyle \displaystyle \begin{align*} x' &= r\cos{( \theta + \alpha )} \\ &= r \left[ \cos{(\theta)}\cos{(\alpha)} - \sin{(\theta)}\sin{(\alpha)} \right] \\ &= r\cos{(\theta)}\cos{(\alpha)} - r\sin{(\theta)}\sin{(\alpha)} \\ &= x\cos{(\alpha)} - y\sin{(\alpha)} \end{align*}

and

\displaystyle \displaystyle \begin{align*} y' &= r\sin{(\theta + \alpha)} \\ &= r\left[ \sin{(\theta)}\cos{(\alpha)} + \cos{(\theta)}\sin{(\alpha)} \right] \\ &= r\sin{(\theta)}\cos{(\alpha)} + r\cos{(\theta)}\sin{(\alpha)} \\ &= y\cos{(\alpha)} + x\sin{(\alpha)} \end{align*}

So putting this all together with the matrices we find

\displaystyle \displaystyle \begin{align*} \left[ \begin{matrix} x' \\ y' \end{matrix} \right] &= \left[ \begin{matrix} x\cos{(\alpha)} - y\sin{(\alpha)} \\ x\sin{(\alpha)} + y\cos{(\alpha)} \end{matrix} \right] \\ \left[ \begin{matrix} x' \\ y' \end{matrix} \right] &= \left[ \begin{matrix} \cos{(\alpha)} & -\sin{(\alpha)} \\ \sin{(\alpha)} & \phantom{-}\cos{(\alpha)} \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] \end{align*}

And so the rotation matrix is \displaystyle \displaystyle \begin{align*} \left[ \begin{matrix} \cos{(\alpha)} & -\sin{(\alpha)} \\ \sin{(\alpha)} & \phantom{-} \cos{(\alpha)} \end{matrix} \right] \end{align*}.

3. ## Re: Proof of the rotation matrix

Konstantine

pare to biblio tis tritis lykeiou mathimatika kateythinsis
ekei tha ta breis sto proto kefalaio prin tous migadikous arithmous.

MINOAS

4. ## Re: Proof of the rotation matrix

Originally Posted by MINOANMAN
Konstantine

pare to biblio tis tritis lykeiou mathimatika kateythinsis
ekei tha ta breis sto proto kefalaio prin tous migadikous arithmous.

MINOAS
Keep it in English folks. PM in whatever language you like.

-Dan

5. ## Re: Proof of the rotation matrix

Hi,
The short answer is $\displaystyle \text{cos}(\theta+\pi/2)=-\text{sin}(\theta)$

For a more detailed explanation:

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