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Question # The coordinates of the point in the x^{prime} y^{prime} - coordinate system with the given angle of rotation and the xy-coordinates.

Alternate coordinate systems
ANSWERED The coordinates of the point in the $$\displaystyle{x}' {y}'$$ - coordinate system with the given angle of rotation and the xy-coordinates. 2020-11-18

Suppose the x- and y- axes are rotated about the origin through a positive acute angle $$\displaystyle\theta$$, then the coordinates $$\displaystyle{\left({x},\ {y}\right)}$$ and $$\displaystyle{\left({x}', {y}'\right)}$$ of a point P in the xy- and $$\displaystyle{x}',{y}'$$- coordinate systems are related by the following formulas:
$$\displaystyle{x}'={x}{\cos{\theta}}\ +\ {y}{\sin{\theta}}$$
$$\displaystyle{y}'=\ -{x}{\sin{\theta}}\ +\ {y}{\cos{\theta}}$$
$$\displaystyle{x}={x}'{\cos{\theta}}\ -\ {y}'{\sin{\theta}}$$
$$\displaystyle{y}={x}'{\sin{\theta}}\ +\ {y}'{\cos{\theta}}$$
Given:
The angle of rotation is $$\displaystyle\theta={30}^{{\circ}}$$ and the x- and y- coordinates are 0 and 2, respectively.
Calculation:
Use the definition, substitute the values of x-, y- coordinates and $$\displaystyle\theta$$ in order to obtain the values of $$\displaystyle{x}',\ {y}'$$ - coordinates.
$$\displaystyle{x}'={\left({0}\right)}{\cos{\ }}{30}^{{\circ}}\ +\ {\left({2}\right)}{\sin{\ }}{30}^{{\circ}}$$
$$\displaystyle{y}'=\ -{\left({0}\right)}{\sin{\ }}{30}^{{\circ}}\ +\ {\left({2}\right)}{\cos{\ }}{30}^{{\circ}}$$
Know that $$\displaystyle{\sin{\ }}{30}^{{\circ}}={\frac{{{1}}}{{{2}}}}\ {\quad\text{and}\quad}\ {30}^{{\circ}}={\frac{{\sqrt{{{3}}}}}{{{2}}}}$$
Thus, the $$\displaystyle{x}'\ {\quad\text{and}\quad}\ {y}'$$ - coordinates become $$\displaystyle{x}'={\left({0}\right)}\ {\left({\frac{{\sqrt{{{3}}}}}{{{2}}}}\right)}\ +\ {\left({2}\right)}{\left({\frac{{{1}}}{{{2}}}}\right)}$$
$$\displaystyle={0}\ +\ {\left({1}\right)}$$
=1
$$\displaystyle{y}'=\ -{\left({0}\right)}{\left({\frac{{{1}}}{{{2}}}}\right)}\ +\ {\left({2}\right)}{\left({\frac{{\sqrt{{{3}}}}}{{{2}}}}\right)}$$
$$\displaystyle={0}\ +\ {\left(\sqrt{{{3}}}\right)}$$
$$\displaystyle=\sqrt{{{3}}}$$
Therefore, the coordinates of the point the $$\displaystyle{x}' {y}'$$ - coordinate system are 1 and $$\displaystyle\sqrt{{{3}}}$$, respectively.