analytic geometrycoordinate systems
I was trying to prove:
To carry out a rotation using matrices the point $(x, y)$ to be rotated from the angle, $θ$, where $(x′, y′)$ are the co-ordinates of the point after rotation, and the formulae for $x′$ and $y′$ can be seen to be
$x'= x \cosθ – y \sinθ$
$y'= x \sinθ + y \cosθ$
But when I prove it by trig/geometry, it has to be split into obtuse and acute case. Is there a way I could go straight forward without casework?
$\newcommand{\Vec}[1]{\mathbf{#1}}$Here is an algorithm (not a formula per se, but can be turned into one):
Geometrically, $(\Vec{e}_{1}', \Vec{e}_{2}', \Vec{e}_{3})$ is the result of rotating the "original" orthonormal basis $(\Vec{e}_{1}, \Vec{e}_{2}, \Vec{e}_{3})$ through angle $\delta$ about $\Vec{e}_{3}$. The formula in item 4. decomposes $A - B$ (the displacement of point $A$ from the "origin" $B$) into components with respect to the original basis, and uses those as components with respect to the rotated basis.
Note: If $\Vec{e}_{3} = (0, 0, \pm 1)$ we can pick $\Vec{e}_{1} = (1, 0, 0)$ and $\Vec{e}_{2} = (0, 1, 0)$. Otherwise, we have $\Vec{e}_{3} = (X, Y, Z)$ with $X^{2} + Y^{2} \neq 0$, and may pick $$ \Vec{e}_{1} = \frac{(-Y, X, 0)}{\sqrt{X^{2} + Y^{2}}}. $$ This is not numerically robust if the segment $\overline{BC}$ is nearly parallel to the third Cartesian axis. In practice, find the two largest (in absolute value) components of $\Vec{e}_{3}$ and pick $\Vec{e}_{1}$ using those instead of the first two.
Best Answer
You can also use polar coordinates. If $(x,y)= (R \cos(\phi), R \sin(\phi))$ then
$$(x',y')=(R \cos( \phi+\theta), R \sin( \phi+\theta)) $$ $$= \left(R\cos(\phi) \cos(\theta) - R\sin(\phi) \sin(\theta) , R\sin(\phi) \cos(\theta) + R\cos(\phi) \sin(\theta) \right) $$ $$= \left(x \cos(\theta) - y \sin(\theta) , x\sin(\theta) + y\cos(\theta) \right) \,.$$
You have to also remeber that this only proves the formula for $(x,y) \neq (0,0)$, so you need to check separatelly that it holds for $(0,0)$ (which is of course trivial)....