Other simple proofs to derive the rotation formulas (for students of an high school)

Question

Given a cartesian reference system that in the diagram I have indicated with capital letters $OXY$ (in black) imagine to rotate it by an angle $α$ counterclockwise (positive angle):

N.B. the name of the axes has been indicated with capital letters, while the coordinates of point $P$ (in the two reference systems) have been indicated with the lower case ones. Given the following construction:

The $hx$ segment is given by:

$$hx = x'\cos(α)$$

while the $oh$ segment is given by:

$$oh = y'\sin(α)$$

The $ox$ segment, which is none other than the $x$ abscissa is given by:

$$x = oh - hx = y'\sin(α) - x'\cos(α)$$

This relationship connects the two reference systems. In the same way we calculate the ordinate $y$ by the following construction:

The $by$ segment is given by:

$$by = x'\sin(α)$$

while the $ob$ segment is given by:

$$ob = y'\cos(α)$$

from which you have:

$$y = ob + by = y'\cos(α) + x'\sin(α)$$

In summary:

$$x = y'\sin(α) - x'\cos(α)$$

$$y = y'\cos(α) + x'\sin(α)$$

This is the proof that I give for my students for the rotation.

Are there other simple and more fast proofs? Everyone that every user to know is welcome.

Answer 1

Depending on your students prerequisite knowledge, there is a simpler way:

1. Establish that rotation is linear.
2. Show that rotation by $\alpha$ counterclockwise rotates the vector $(1,0)$ to $(\cos\alpha,\sin\alpha)$ and $(0,1)$ to $(-\sin\alpha,\cos\alpha)$. (similar to what you did, but simpler to draw and understand, imho)
3. Use this to find the matrix representation $\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$

Keep in mind, though, that I described how to find the formula for an active transformation, that is, how to rotate a vector within the given coordinate system. What you did was a passive transformation, rotating the coordinate system. The difference is just that to find the coordinates after a passive transformation, you have to apply the inverse of the above matrix instead.

Answer 2

You can use polar coordinates and the angle addition formulas. The case $(x,y)=(0,0)$ is trivial, now for $(x,y)\neq(0,0),$ let $(x,y)=(r\cos(\theta),r\sin(\theta)).$

Then $$(x',y')=(r\cos(\theta+\alpha),r\sin(\theta+\alpha))$$ $$=(r\cos(\theta)\cos(\alpha)-r\sin(\theta)\sin(\alpha),r\sin(\theta)\cos(\alpha)+r\cos(\theta)\sin(\alpha))$$ $$=(x\cos(\alpha)-y\sin(\alpha),x\sin(\alpha)+y\cos(\alpha)).$$

Answer 3

Use scalar prodact $$ \mathbf{r} = x\ \mathbf{\hat x} +y\ \mathbf{\hat y} = x'\ \mathbf{\hat x}' +y'\mathbf{\hat y}' $$ $$ \mathbf{\hat x}'\cdot \mathbf{\hat x}'=1,\qquad \mathbf{\hat y}'\cdot \mathbf{\hat y}'=1,\qquad \mathbf{\hat x}'\cdot \mathbf{\hat y}'=\mathbf{\hat y}'\cdot \mathbf{\hat x}'=0 $$ $$ \mathbf{r}\cdot \mathbf{\hat x}' = x' = x\ \mathbf{\hat x}\cdot \mathbf{\hat x}' +y\ \mathbf{\hat y}\cdot \mathbf{\hat x}'\\ \mathbf{r}\cdot \mathbf{\hat y}' = y' = x\ \mathbf{\hat x}\cdot \mathbf{\hat y}' +y\ \mathbf{\hat y}\cdot \mathbf{\hat y}' $$ The rotation is considered from $\mathbf{\hat x}$ to $\mathbf{\hat x}'$ by an angle $\alpha$ $$ \mathbf{\hat x}\cdot \mathbf{\hat x}'=\cos \alpha,\qquad \mathbf{\hat y}\cdot \mathbf{\hat y}'=\cos \alpha,\qquad \mathbf{\hat x}\cdot \mathbf{\hat y}'= \cos (\alpha+\frac{\pi}{2}) \qquad \mathbf{\hat y}\cdot \mathbf{\hat x}'= \cos (\alpha-\frac{\pi}{2}) $$ Recall that $$ \cos(\alpha+\beta) = \cos\alpha\cos\beta-\sin\alpha\sin\beta $$ that means $$ \cos (\alpha-\frac{\pi}{2})= +\sin\alpha \qquad \cos (\alpha-\frac{\pi}{2})=-\sin\alpha $$

Answer 4

I have thought to this alternative solution:

Let $P\equiv(x,y)$ in the reference system $xOy$ and let $$\vec{OP}=x\hat{\bf x}+y\hat{\bf y}\equiv (x,y)$$ the displacement vector. In the reference system $XOY$ let, $$\vec{OP}=X\hat{\bf X}+Y\hat{\bf Y}\equiv (X,Y)$$

Hence: $$\vec{OP}=x\hat{\bf x}+y\hat{\bf y}=X\hat{\bf X}+Y\hat{\bf Y}\tag 1$$ We suppose to rotate in anticlockwise the old reference system $xOy$ of an angle $\theta$. I play with the colours (red vectors and dotted lines) of the triangles generate by $\hat{\bf X}$ and $\hat{\bf Y}$.

We have hence:

$$\hat{\bf X}=(\cos\theta,\sin\theta)=\cos\theta\hat{\bf x}+\sin\theta\hat{\bf y}$$ taking the right square triangle for $\hat{\bf X}$. Taking the left triangle for $\hat{\bf Y}$ comparing the lenghts, it is evident that:

$$\hat{\bf Y}=(-\sin \theta,\cos \theta)=-\sin\theta\hat{\bf x}+\cos\theta\hat{\bf y}$$

Definitively from the $(1)$,

$$\vec{OP}=X\hat{\bf X}+Y\hat{\bf Y}=X(\cos\theta\hat{\bf x}+\sin\theta\hat{\bf y})+Y(-\sin\theta\hat{\bf x}+\cos\theta\hat{\bf y})$$ and being that

$$x\hat{\bf x}+y\hat{\bf y}=\vec{OP}\equiv (X\cos\theta-Y\sin\theta)\hat{\bf x}+(X\sin\theta+Y\cos\theta)\hat{\bf y}$$ we have, definitively:

\begin{cases} x=X\cos\theta-Y\sin\theta &\\ y=X\sin\theta+Y\cos\theta \end{cases}