Transformation of axes by rotation
Matthew Harrington 
How can I intutively understand the formula for getting new coordinate of point P after rotation of axes which was P(x,y) with respect to the old axes?
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$\begingroup$Consider initially that the vectors are written in the canonical base, so rotating this vector is equivalent to rotating the canonical base and then seeing the coordinates of that vector in this new base, looking at the figure that rotates the base the vectors of the canonical base in a theta angle, we can see that
$$e'_1=\left(\cos \left(θ\right),\sin\left(θ\right)\right)$$ and $$e'_2=\left(-\sin\left(θ\right),\cos \left(θ\right)\right)$$
then the transformation is the matrix
$$ M= \left[ \begin{array}{cc} \cos(\theta) & \sin(\theta)\\ -\sin(\theta) & \cos(\theta)\\ \end{array} \right]$$
so after rotating the vector $$\left(x,y\right)$$ we will have
$$\left(x\cos \left(θ\right)+y\sin\left(θ\right),-x\sin\left(θ\right)+y\cos \left(θ\right)\right)$$
$\endgroup$ $\begingroup$Consider the following figure
Here the red coordinate system is rotated by $\alpha$.
We have for the black coordinates $$x=R\cos(\beta),y=R\sin(\beta)$$,
Also, for the red coordinates we have
$$x'=R\cos(\gamma), y'=R\sin(\gamma).$$
Notice that $\gamma=\beta-\alpha$ and use the adequate trig formula:
$$x'=R\cos(\beta-\alpha)=R(\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta))=x\cos(\alpha)+y\sin(\alpha)$$
and ...
$\endgroup$ $\begingroup$Let old point be at $1$ and new point be at $2$ both in first quadrant. Let the rotation angle be a small $\theta$ counter-clockwise.
$$x_{new} =x_{old} \, cos \,\theta -y_{old} \,sin \,\theta \tag1 $$
$$y_{new} = x_{old} \,\sin\theta + y _{old}\cos\theta \tag2 $$
It is noticed that
as far as new $x$ is concered there is a minus sign at right hand side and its x-coordinate got reduced.
as far as new $y$ is concered there is a plus sign at right hand side and its y-coordinate got increased.
It's hard to see what you mean by intuitive understanding when talking about a system of equations. But here's some context that may help more.
As you must know the formulas
$$\begin{array} &x=x'\cos\phi& -&y'\sin\phi&\\ y=x'\sin\phi &+&y'\cos\phi& \end{array}$$
give us a way to transform an equation $F(x,y)=0$ into $G(x',y')=0.$ This gives a map (a rotation as you already know) from the system of coordinates $(x,y)$ into $(x',y').$ Solving explicitly for $x',y'$ in the system of equations gives
$$\begin{array} &x'=x\cos\phi &+&y\sin\phi\\ y'=-x\sin\phi &+&y\cos\phi \end{array},$$
and the last system may be written in matrix form as $v=Mu,$ where $v=(x',y')^T,\,u=(x,y)^T$ and
$$\begin{pmatrix} \cos\phi &\sin\phi\\ -\sin\phi &\cos\phi \end{pmatrix}.$$
This is a linear transformation of $\mathrm R^2.$ To see that it is a rotation, compute the determinant to see that it is $1.$ Thus, the squares in the rectangular system $(x,y)$ preserve their orthogonality, so that their corresponding area in the transform system $(x',y')$ remains the same.
$\endgroup$ $\begingroup$Assume the ray to the point (x,y) from the origin makes an angle of $\alpha$ anti-clockwise with +ve x-axis.
Then (x,y) = (r cos$\alpha$,r sin$\alpha$) where $r = \sqrt{x^2+y^2}$.
If the point is rotated by an angle of $\theta$ anti-clockwise around the origin, then the point makes an angle of $\alpha+\theta$ with the +ve x-axis. Since the distance of the point from the origin doesn't change due to the rotation around it, the new co-ordinates will be $(X,Y) = (r \cos(\alpha+\theta),r \sin(\alpha+\theta))$.
Using basic trigonometric identities,
$(X,Y)=(r \cos(\alpha+\theta),r \sin(\alpha+\theta))= \\ =(r \cos\alpha \cos\theta-r \sin\alpha\sin\theta, r \cos\alpha\sin\theta-r \sin\alpha \cos\theta) = \\ = (x \cos\theta-y \sin\theta,x \sin\theta+y \cos\theta)$
P.S: Can a kind soul format my answer better (Not familiar with MathJax). Thanks.
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