Problem 41 - Algebra and Trigonometry

Question

If there is a \(60^{\circ}\) angle from the positive \(x\) -axis to the positive \(x^{\prime}\) -axis, explain how to obtain the rotation formulas for \(x\) and \(y\)

Step-by-Step Solution

Answer

The rotation formulas for coordinates x and y when there's a \(60^{\circ}\) rotation from the positive \(x\) -axis to the positive \(x^{\prime}\) -axis are \(x^{\prime} = 1/2x - \sqrt{3}/2y\) and \(y^{\prime} = \sqrt{3}/2x + 1/2y\).

Step 1: Understand the rotation matrix

The rotation matrix for a counterclockwise rotation by an angle \(\theta\) is \(\begin{bmatrix} cos(\theta) & -sin(\theta) \ sin(\theta) & cos(\theta) \end{bmatrix}\). This matrix is derived from the unit circle and basic trigonometry.

Step 2: Substitute the given rotation angle in the rotation matrix

Substitute the given angle of \(60^{\circ}\) into the rotation matrix. Note that this should be converted to radians because the trigonometric functions in the matrix are in terms of radians. So, \(60^{\circ}\) is equivalent to \(\pi/3\) radians. The substitution yields \(\begin{bmatrix} cos(\pi/3) & -sin(\pi/3) \ sin(\pi/3) & cos(\pi/3) \end{bmatrix} = \begin{bmatrix} 1/2 & -\sqrt{3}/2 \ \sqrt{3}/2 & 1/2 \end{bmatrix}\).

Step 3: Apply the rotation matrix to the original coordinates

To find the new transformed coordinates \((x^{\prime}, y^{\prime})\), multiply the rotation matrix by the original coordinate vector \((x, y)\). This yields \(\begin{bmatrix} 1/2 & -\sqrt{3}/2 \ \sqrt{3}/2 & 1/2 \end{bmatrix} \times \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 1/2x - \sqrt{3}/2y \ \sqrt{3}/2x + 1/2y \end{bmatrix}\).