# Rotation of 2 X 2 Cosine and Sine Equations

Given the set of two equations (transformation equations):

$$\bar{x} = x \cos{\alpha} + y \sin{\alpha}$$

$$\bar{y} = -x \sin{\alpha} + y \cos{\alpha}$$

What would be the steps to obtain the inverse transformation equations:

$$x = \bar{x} \cos{\alpha} - \bar{y} \sin{\alpha}$$

$$y= \bar{x} \sin{\alpha} + \bar{y} \cos{\alpha}$$

Question: These Cosine and Sine equations look like a $2 \times 2$ matrix but they are equations. Can one still use the same procedure as when finding the inverse of a $2 \times 2$ matrix? Also, can someone include detailed steps to go from one set of equations to the other set? Thank you.

• Perhaps you could apply a rotation matrix to your image before actually posting it. – user296602 Nov 1 '17 at 3:50
• $\cos\alpha$ and $\sin\alpha$ are both constants in these equations. – amd Nov 1 '17 at 3:56
• The 2nd set of equations are the inverses of the 1st set of equations. That is easy. But, What is the property or rule to change the left side of the equations from X- and Y- bar to the X and Y? – Salvador Perez Nov 1 '17 at 4:06

hint:$$\quad{ \bar{x} = x \cos{\alpha} + y \sin{\alpha}\\\bar{y} = -x \sin{\alpha} + y \cos{\alpha} \\\begin{bmatrix}\bar{x} \\\bar{y} \end{bmatrix}=\begin{bmatrix}\cos{\alpha} & \sin{\alpha} \\-\sin{\alpha} & \cos{\alpha} \end{bmatrix}\begin{bmatrix}x \\y \end{bmatrix}\\\to \begin{bmatrix}\bar{x} \\\bar{y} \end{bmatrix}=R\begin{bmatrix}x \\y \end{bmatrix}\to \\R^{-1}\begin{bmatrix}\bar{x} \\\bar{y} \end{bmatrix}=R^{-1}R\begin{bmatrix}x \\y \end{bmatrix}\\\begin{bmatrix}x \\y \end{bmatrix}=R^{-1}\begin{bmatrix}\bar{x} \\\bar{y} \end{bmatrix}\\\begin{bmatrix}x \\y \end{bmatrix}=\frac{1}{det(R)}\begin{bmatrix}\cos{\alpha} & -\sin{\alpha} \\+\sin{\alpha} & \cos{\alpha} \end{bmatrix}\begin{bmatrix}\bar{x} \\\bar{y} \end{bmatrix} }$$ $remark:det(R)=\sin^2 \alpha+\cos^2 \alpha=1$