Given a point on a unit circle with coordinates $(x, y)$ and angle $ heta$ from the positive $x$-axis, find the coordinates of the point after rotating by $45^circ$ counterclockwise.
Answer 1
Given the initial coordinates $(x, y)$ and angle $\theta$, the coordinates after rotating by $45^\circ$ counterclockwise can be found using the rotation matrix:
$ \begin{bmatrix} \cos(45^\circ) & -\sin(45^\circ) \\ \sin(45^\circ) & \cos(45^\circ) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $
Since $\cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2}$, the formula becomes:
$ \begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $
Performing the matrix multiplication, we get:
$ \begin{bmatrix} \frac{\sqrt{2}}{2}x – \frac{\sqrt{2}}{2}y \\ \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2}y \end{bmatrix} $
Thus, the new coordinates are:
$ \left( \frac{\sqrt{2}}{2}(x – y), \frac{\sqrt{2}}{2}(x + y) \right) $
Answer 2
To find the new coordinates after a $45^circ$ counterclockwise rotation, use the rotation formula:
$ x’ = x cos(45^circ) – y sin(45^circ) $
$ y’ = x sin(45^circ) + y cos(45^circ) $
Since $cos(45^circ) = sin(45^circ) = frac{sqrt{2}}{2}$, substitute these values in:
$ x’ = x cdot frac{sqrt{2}}{2} – y cdot frac{sqrt{2}}{2} $
$ y’ = x cdot frac{sqrt{2}}{2} + y cdot frac{sqrt{2}}{2} $
Simplify to get the new coordinates:
$ x’ = frac{sqrt{2}}{2}(x – y) $
$ y’ = frac{sqrt{2}}{2}(x + y) $
So, the new coordinates after rotation are:
$ left( frac{sqrt{2}}{2}(x – y), frac{sqrt{2}}{2}(x + y)
ight) $
Answer 3
The new coordinates after a $45^circ$ counterclockwise rotation are:
$ x’ = frac{sqrt{2}}{2}(x – y) $
$ y’ = frac{sqrt{2}}{2}(x + y) $
So the final coordinates are:
$ left( frac{sqrt{2}}{2}(x – y), frac{sqrt{2}}{2}(x + y)
ight) $
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