Rotation through any angle θ

Question

They say that rotation of any point $(x,y)$ through any angle $\theta$ is given by $(x \cos\theta, y \sin\theta)$. Can anybody tell how was this derived? Please post here or send me by email.

Answer 1

Since rotations of $\Bbb R^2$ are $\Bbb R$ linear transformations, they can be described by $2\times 2$ matrices over $\Bbb R$. Furthermore, the action of the transformation is completely determined by what it does to the unit vectors $[1,0]$ and $[0,1]$.

In the following, you can take $\theta\in [0,\pi/2]$ as an illustration, although it is valid for any $\theta$.

Taking the line segment between $[0,0]$ and $[1,0]$ and rotating it by $\theta$ radians you get a slanted line segment. Drop a perpendicular to the $x$-axis, and you have a right triangle whose angle on the origin is $\theta$. Clearly the coordinates that $[1,0]$ was rotated to is $[\cos(\theta),\sin(\theta)]$.

Similarly, look at where the line segment between $[0,0]$ and $[0,1]$ goes. It will be a slanted line in the third quadrant (for our "small" theta), and it makes an angle of $\theta$ with the $y$ axis. Dropping a perpendicular towards the $y$-axis and performing a little trigonometry you get that the coordinates of $[0,1]$ must be $[-\sin(\theta),\cos(\theta)]$.

So: what would the transformation matrix $A$ have to look like?

Well, we know that $[1,0]A=[\cos(\theta),\sin(\theta)]$ and that $[0,1]A=[-\sin(\theta),\cos(\theta)]$, so that tells us what the rows of $A$ are:

$$ A=\begin{bmatrix}\cos(\theta)&\sin(\theta)\\ -\sin(\theta)&\cos(\theta)\end{bmatrix} $$

Finally, then we know that this formula applies to anything in the span of $[1,0]$ and $[0,1]$, not just those two vectors. Rotating $[x,y]$ is acheived by matrix multiplication:

$$ [x,y]A=[x,y]\begin{bmatrix}\cos(\theta)&\sin(\theta)\\ -\sin(\theta)&\cos(\theta)\end{bmatrix}=[x\cos(\theta)-y\sin(\theta),x\sin(\theta)+y\cos(\theta)] $$

Answer 2

I want to provide a very interesting method. Based on Euler equation:

$$e^{iθ} = \cosθ+i \sinθ$$

it can be taken as an "rotation" action. So, if the original coordinate of your point is $(x,y)$, it can be presented by $x + iy$. Then, the coordinate after rotation is going to be

$$(x + iy) e^{iθ}$$

Then, you get

$$[x\cosθ - y\sinθ] + i [x\sinθ + y\cosθ]$$

Answer 3

I arrived at the end formula $[x\cdot\cos(\theta)−y\cdot\sin(\theta),x\cdot\sin(\theta)+y\cdot\cos(\theta)]$ using the unit circle, multiplied by some radius $r$. Any point on the circle is $(r\cdot\cos\theta, r\cdot\sin\theta)$. Rotate further through any other angle A and the new coordinates are $(r\cdot\cos(\theta+A), r\cdot\sin(\theta+A))$.

Using the formulas for $\sin(u+v)$ and $\cos(u+v)$ you can expand those expressions. Substitute $x$ for $r\cdot\cos \theta$ and $y$ for $r\cdot\sin \theta$ and you'll get the same thing (in terms of $A$, the angle rotated through).