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I have seen in diferent math posts that the following matrix: $$\begin{pmatrix}\cos t&-\sin t\\\sin t&\cos t\end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix}$$ rotates a point $p$ in the plane counterclockwise by an angle of $t$. I was wondering if someone could explain why this matrix works as it does.


I'm not even sure if this is correct, nor if it relates to the matrix I'm trying to understand, but this is what I have tried:

Suppose you want to rotate the point $p = (p_x,p_y)$ by an angle of $t$. Let's call this new point $q = (q_x,q_y)$.

$q_x = p_x cos (t) + p_x sin (t)$,

$q_y= p_y cos (t+π/2) + p_y sin(t+π/2)$

and we have

$$\begin{pmatrix}\ p_x cos (t)&\ p_x sin (t)\\\ p_y sin (t+π/2)&\ p_ycos (t+π/2)\end{pmatrix}=\begin{pmatrix}q_x \\q_y\end{pmatrix}$$

With $q_x + q_y = q$


Any thoughts/ideas would be really appreciated.

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  • $\begingroup$ Have you tried visualizing this on a graph? Take a point on a graph $(x, y)$ and see what happens to it. Alternatively, take any point $(x, y)$ and see what you need to do to it to rotate it anticlockwise by some angle $t$ $\endgroup$ – Jihoon Kang Aug 19 '17 at 8:42
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Way 1

$$\begin{pmatrix}X \\ Y\end{pmatrix}=\begin{pmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}\implies \begin{cases}X=\cos(t)x-\sin(t) y\\Y=\sin(t)x+\cos(t)y\end{cases}.$$

Now, one can show that $$X^2+Y^2=x^2+y^2,$$ what mean that $(x,y)$ and $(X,Y)$ belong to the same circle. Therefore, $(X,Y)$ is the image of $(x,y)$ by a rotation of center $(0,0)$. To find the angle, one can compute $$\frac{\left<\begin{pmatrix}x\\ y\end{pmatrix},\begin{pmatrix}X\\ Y\end{pmatrix}\right>}{\left\|\begin{pmatrix}x\\ y\end{pmatrix}\right\|\left\|\begin{pmatrix}X\\ Y\end{pmatrix}\right\|},$$ what gives the cosinus between $\begin{pmatrix}x\\ y\end{pmatrix}$ and $\begin{pmatrix}X\\ Y\end{pmatrix}$ and conclude.

Way 2

If you know complex number, you can prove that $$X+iY=\sqrt{x^2+y^2}e^{it}=\sqrt{x^2+y^2}(\cos(t)+i\sin(t)),$$ and it allow you to conclude.

Added

$$\frac{\left<\begin{pmatrix}x\\ y\end{pmatrix},\begin{pmatrix}X\\ Y\end{pmatrix}\right>}{\left\|\begin{pmatrix}x\\ y\end{pmatrix}\right\|\left\|\begin{pmatrix}X\\ Y\end{pmatrix}\right\|}=\frac{\cos(t)x^2-xy\sin(t)+xy\sin(t)+\cos(t)y^2}{x^2+y^2}=\frac{x^2+y^2}{x^2+y^2}\cos(t)=\cos(t).$$

Therefore the angle between $(x,y)$ and $(X,Y)$ is $t$.

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  • $\begingroup$ I really don't know much about math. In "Way 1" I understand everything up to $X^2+Y^2=x^2+y^2$, however, I don't know the meaning of that weird thing in those weird parenthesis, could you at least tell me how is it called so that I can google it? Thank for your answer by the way. $\endgroup$ – Leo Aug 19 '17 at 9:14
  • $\begingroup$ $\left<\begin{pmatrix}X\\ Y\end{pmatrix},\begin{pmatrix}y\\ y\end{pmatrix}\right>=Xx+Yy$ is the scalar product, and $\left\|\begin{pmatrix}a\\ b\end{pmatrix}\right\|=\sqrt{a^2+b^2}$ is the norm. $\endgroup$ – Surb Aug 19 '17 at 9:17
  • $\begingroup$ I'm truly sorry that I don't understand, but could you show the complete proof of "Way 1"? $\endgroup$ – Leo Aug 19 '17 at 9:30
  • $\begingroup$ @Leo : I edit my answer. But it's probably not the way you have to do if you don't know all those notion. Maybe a draw would be the best for you... $\endgroup$ – Surb Aug 19 '17 at 9:57

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