- Using " matrix by vector" multiplication, one can establish that the coordinates of the image ($P'$) of a point $P=(a,b)$ under a rotation by $\alpha$ radians ( counterclockwise) are :
$$ \Bigg(a\space cos(\alpha) - b\space sin (\alpha) , a\space sin (\alpha) + b\space cos (\alpha)\Bigg)$$
- One can also find an expression using the method consisting in considering the rotated point $P'$ as the result of a rotation (counterclockwise) of the original point $P=(a,b)$ on a circle of radius $D = \sqrt {a^2 + b^2}$ from an initial angle $I= \arctan( b/a)$. Thus we obtain
$$P' = \Bigg( D \space cos(I+\alpha) , D \space sin(I+\alpha)\Bigg)$$
$$= \Bigg (\sqrt {a^2 + b^2} \space cos( (\arctan \frac{b} {a}) + \alpha) , \sqrt {a^2 + b^2} \space sin( (\arctan \frac{b} {a}) + \alpha) \Bigg )$$
- From this follows that :
$$ a\space cos(\alpha) - b\space sin (\alpha) = \sqrt {a^2 + b^2} \space cos( (\arctan \frac{b} {a}) + \alpha)$$
and
$$ a\space sin (\alpha) + b\space cos (\alpha) = \sqrt {a^2 + b^2} \space sin( (\arctan \frac{b} {a}) + \alpha)$$
I suppose that, in both cases, no-one is interested in rewriting the LHS as the RHS, but is it also true the other way around?