shifting coordinate system in 2D...solving simultaneous equation

[SUDOnym]

When in 2D, the coordinates of a place in space vary depending on the coordinate axes that are being used given by:

A_{x}^{\prime}=A_{x}\cos\theta+A_{y}\sin\theta (1)

and

A_{y}^{\prime}=-A_{x}\sin\theta+A_{y}\cos\theta (2)

Now I am trying to reverse it - to show what A_x and A_y are in terms of A_x' and A_y'..This is a simple simultaneous eqtn but I cannot get to the result that I know is the correct answer, the following is the relationship I wish to show:

A_{x}=A_{x}^{\prime}\cos\theta-A_{y}^{\prime}\sin\theta (3)


My attempt:

firstly, rearrange equation (1) to get:

A_{y}=\frac{-A_{x}\cos\theta+A_{x}^{\prime}}{\sin\theta} (4)

now plug (4) into (2) to eliminate A_y:

A_{y}^{\prime}=-A_{x}\sin\theta+A_{x}^{\prime}\arctan\theta-A_{x}\cos\theta\arctan\theta (5)

(From now on I stop putting in the thetas for speed..)

\implies A_{y}^{\prime}\tan=-A_{x}\sin\tan-A_{x}\cos+A_{x}^{\prime} (6)

\implies\frac{A_{y}^{\prime}\tan-A_{x}^{\prime}}{cos}=-A_{x}\tan^{2}-A_{x} (7)


I cannot not see how to progress after this point...I don't know if I have made a mistake in the algebra (although I have checked a few times) or if I am missing a useful trig identity.

[Ray Vickson]

You could use Cramer's Rule to solve for A_x and A_y in terms of A'_x and A'_y. In Cramer's Rule, you need to divide through by the determinant of the coefficient matrix. What is that determinant in your case?

RGV

[vela]

When in 2D, the coordinates of a place in space vary depending on the coordinate axes that are being used given by:

A_{x}^{\prime}=A_{x}\cos\theta+A_{y}\sin\theta (1)

and

A_{y}^{\prime}=-A_{x}\sin\theta+A_{y}\cos\theta (2)

Now I am trying to reverse it - to show what A_x and A_y are in terms of A_x' and A_y'..This is a simple simultaneous eqtn but I cannot get to the result that I know is the correct answer, the following is the relationship I wish to show:

A_{x}=A_{x}^{\prime}\cos\theta-A_{y}^{\prime}\sin\theta (3)


My attempt:

firstly, rearrange equation (1) to get:

A_{y}=\frac{-A_{x}\cos\theta+A_{x}^{\prime}}{\sin\theta} (4)
Fine up to here, but then...
now plug (4) into (2) to eliminate A_y:

A_{y}^{\prime}=-A_{x}\sin\theta+A_{x}^{\prime}\arctan\theta-A_{x}\cos\theta\arctan\theta (5)
This is wrong. \cos\theta/\sin\theta does not equal \arctan\theta. Notation-wise, this is how it works:
\begin{align}
\tan^{-1} x & \Leftrightarrow \arctan x \\
(\tan x)^{-1} & \Leftrightarrow 1/\tan x
\end{align}
It's confusing because \tan^2 x does mean (\tan x)^2. It's just one of those inconsistencies in common notation you have to learn and be aware of.

In this problem, you might as well leave everything in terms of sine and cosine since that's what you want in the end anyway.
(From now on I stop putting in the thetas for speed..)

\implies A_{y}^{\prime}\tan=-A_{x}\sin\tan-A_{x}\cos+A_{x}^{\prime} (6)

\implies\frac{A_{y}^{\prime}\tan-A_{x}^{\prime}}{cos}=-A_{x}\tan^{2}-A_{x} (7)


I cannot not see how to progress after this point...I don't know if I have made a mistake in the algebra (although I have checked a few times) or if I am missing a useful trig identity.
If you factor the RHS of equation (7), you get -(\tan^2\theta+1)A_x. You can simplify that using a trig identity.

[epenguin]

Not clear how you know the thing you wish to show, but it is usually an advantage in a math problem to know the answer.

I think the suggested methods are worth working through. However surely simpler wouldn't it be to say: you already have the formulae for the new co-ordinates (A'x, A'y) of a point whose co-ordinates with respect to the previous co-ordinates were (Ax. Ay) after rotation through θ.

This pair of formulae are general for any 2-D rotation.

So now what is it you have to find? The formulae for the new co-ordinates (Ax, Ay)of a point whose co-ordinates with respect to the previous co-ordinates were (A'x. A'y) after rotation through -θ. So apply your general formula and it is pretty immediate.