Rotating the axes and changing variables

Question

Without using matrix inversion, what is the easiest way to show that, for variables $u$ and $v$ corresponding to rotating the $x-y$ axes by $\theta$, we have $$ x=u\cos \theta -v \sin \theta\\ y=u\sin\theta+v\cos\theta $$ My attempt: Using polar coordinates $x=r\cos\alpha$ and $y=r\sin\alpha$, we get $$ u=r\cos(\alpha-\theta)=r\cos\alpha\cos\theta+r\sin\alpha\sin\theta\\ v=r\sin(\alpha-\theta)=r\sin\alpha\cos\theta+r\cos\alpha\sin\theta $$ Without using matrix inversion, how can we invert this? I guess it is ultimately matrix inversion in disguise, unless there is a more natural way of doing it. The reason I ask is that my students have not yet dealt with matrices, which is unfortunate.

Answer 1

One way is to include basis vectors. The same vector can be described in two different coordinate systems just with different components and basis vectors (passive transformation). $$\vec x = x \: \hat{e_x} + y \: \hat{e_y} = u \: \hat{e_u} + v \: \hat{e_v}$$

The rotated basis vectors are clearly given by $$\hat{e_u} = \hat{e_x} cos(\theta) + \hat{e_y} sin(\theta)$$ $$\hat{e_v} = - \hat{e_x} sin(\theta) + \hat{e_y} cos(\theta)$$ Substitute these back in $\vec x$ $$\vec x = u (\hat{e_x} cos(\theta) + \hat{e_y} sin(\theta)) + v ( - \hat{e_x} sin(\theta) + \hat{e_y} cos(\theta)) $$ $$= (u \: cos(\theta) - v \: sin(\theta)) \hat{e_x} + (u \: sin(\theta) + v \: cos(\theta)) \hat{e_y} $$ And comparing with $\vec x = x \hat{e_x} + y \hat{e_y}$ component-wise gives $$x =u \: cos(\theta) - v \: sin(\theta)$$ $$y = u \: sin(\theta) + v \: cos(\theta)$$

Answer 2

Like you have mentioned $u = r\cos(\alpha -\theta)$ and $v = r\sin(\alpha-\theta)$.

For the first equation multiply $2\cos\theta$ on both sides. We get $2u\cos\theta = r^2\cos\theta\cos(\alpha-\theta)$ which can be written as $2u\cos\theta = r(\cos\alpha + \cos(\alpha - 2\theta)$.

This implies $2u\cos\theta = x + r\cos(\alpha -2\theta)$.

Similarly multiply $2\sin\theta$ on the second equation to get $2v\sin\theta = r\cos(\alpha -2\theta) - x$.

Subtracting we get $2u\cos\theta - 2v\sin\theta = 2x$ which implies $x = u\cos\theta - v\sin\theta$.

Value of $y$ can also be found in a similar fashion by this time multiplying $\sin\theta$ initially and then $\cos\theta$.

Hope this helps!