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

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

and

[tex]A_{y}^{\prime}=-A_{x}\sin\theta+A_{y}\cos\theta[/tex] (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:

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

My attempt:

firstly, rearrange equation (1) to get:

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

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

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

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

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

[tex]\implies\frac{A_{y}^{\prime}\tan-A_{x}^{\prime}}{cos}=-A_{x}\tan^{2}-A_{x}[/tex] (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.

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

and

[tex]A_{y}^{\prime}=-A_{x}\sin\theta+A_{y}\cos\theta[/tex] (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:

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

My attempt:

firstly, rearrange equation (1) to get:

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

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

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

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

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

[tex]\implies\frac{A_{y}^{\prime}\tan-A_{x}^{\prime}}{cos}=-A_{x}\tan^{2}-A_{x}[/tex] (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.

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