Rotations alternative proofs?

Question

Let $P(x,y)$ be any point and let $\rho$ and $\alpha$ be its polar coordinates. We have \begin{align*} x &= \rho\cos\alpha \\ y &= \rho\sin\alpha \end{align*} the point $P'(x',y')$, image of $P$ in a counterclockwise rotation of an angle $\theta$, has polar coordinates $(\rho,\alpha+\theta)$. Its Cartesian coordinates are therefore given by the preceding system, where we put $\alpha+\theta$ in place of $\alpha$: \begin{align*} x' &= \rho\cos\left(\alpha + \theta\right) \\ y' &= \rho\sin\left(\alpha + \theta\right). \end{align*} applying the addition formulas of sine and cosine and also taking into account the initial formulas, we obtain the rotation formulas, in fact: \begin{align*} x' &= \rho\cos \left(\alpha+\theta\right) = \rho\left(\cos \alpha \cos \theta - \sin\alpha\sin\theta\right) = x\cos\theta-y\sin\theta, \\ y' &= \rho\sin\left(\alpha+\theta\right) =\rho\left(\sin \alpha \cos \theta + \cos \alpha\sin\theta\right) = x\sin\theta+y\cos\theta. \end{align*} Although it is simple it is necessary to remember goniometry in the third year of high school, which is not contemplated. Is there an alternative proof?

Answer 1

We define $\phi:\mathbb{R^2}\rightarrow \mathbb{R^2}$ to be the rotation about the origin (0,0) by an angle $\theta$. We first will show that this is a linear map. We note that a rotation sends a parallelogram to a parallelogram( You can check this on your own). Let $x,y\in \mathbb{R^2}$, draw the line segment between x and the origin and do the same for y. Now you have two line segments, draw another two such that this becomes a parallelogram which we call T=(0,x,y,p) with x+y=p. If we rotate x by an angle $\theta $ and y too by an angle of $\theta$, then we have rotated our parallelogram T by an angle of $\theta$, which means every point in T is rotated by an angle of $\theta $. In particular we have that p is rotated by $\theta$. It follows that $\phi(x)+\phi(y)=\phi(p)=\phi(x+y)$. So rotating x+y=p by an angle $\theta$ is the same as first rotating x by an angle of $\theta$ and than y by an angle $\theta$ so this map is additive.

In a similar fashion you can multiply x by a constant $c\in \mathbb{R}$ and then rotate it by $\theta$. This is the same as first rotating x by an angle $\theta $ and then multiplying it by c. So we have $\phi(cx)=c\phi(x)$

We just proved that $\phi$ is linear.

Now what happens if we want to rotate (1,0) and (0,1). Let's start with (1,0): (1,0) is in the unit circle, so a rotation by an angle of $\theta $ of this point will still remain on the unit circle, since a rotation is an isometry and preserves lengths. The x and y coordinate of the rotated point are than $(\cos(\theta),\sin(\theta)) $, you can verify this by drawing a triangle :

So we have $\phi((1,0))=(\cos(\theta),\sin(\theta))$

In the same way we can do this for (0,1) : Draw a circle and rotate the point (0,1) by a clockwise rotation of an angle $\theta$. After you did that try to find its respective coordinates after the rotation(you can draw a triangle again). What you will get is : $\phi((0,1))=(-\sin(\theta ),\cos(\theta))$.

You may think: Why are the images of (1,0) and (0,1) important?. Given any point (x,y) and using the fact that the map $\phi$ is linear we have the following : $$\phi((x,y))=\phi((x,0)+(0,y))=\phi(x,0)+\phi(0,y)=\phi(x\cdot1,0)+\phi(0,y\cdot1)=x\phi(1,0)+y\phi(0,1)$$$$=(x\cos(\theta),x\sin(\theta))+(-y\sin(\theta),y\cos(\theta))=(x\cos(\theta)-y\sin(\theta),x\sin(\theta)+y\cos(\theta))$$