How do you rotate a figure 60 degrees clockwise about origin?

Answer 1

Suppose there is another x'y'-coordinate system that has the same origin as the xy-coordinate system, and θ is the angle from the positive x-axis to the positive x'-axis. If there is a point (x, y) in the xy-coordinate system, and a point (x', y') in the rotated x'y'- coordinate system, then

x = x' cos θ - y' sin θ and

y = x' sin θ + y' cos θ (rotation of axis formulas)

Since the rotation of 60 degrees clockwise, is the same as the rotation of 300 degrees anticlockwise,

then cos 300ᵒ = cos (-60ᵒ) = 1/2 and sin 300ᵒ = sin (-60ᵒ) = -√3/2 (only cosine is positive in the IV quadrant).

So we need to express x' and y' in terms of x and y.

x = x' cos θ - y' sin θ

x = (1/2)x' - (-√3/2)y' multiply by 2 each term to both sides

2x = x' + (√3)y' subtract (√3)y' to both sides

x' = 2x - (√3)y'

y = x' sin θ + y' cos θ

y = (-√3/2)x' + (1/2)y' multiply by 2 to both sides

2y = (-√3)x' + y' add (√3)x' to both sides

y' = (√3)x' + 2y

so that,

x' = 2x - (√3)y' replace y' by (√3)x' + 2y

x' = 2x - √3[(√3)x' + 2y]

x' = 2x - 3x' - 2√3y add 3x' to both sides

4x' = 2x - 2√3y divide by 4 to both sides

x' = (1/2)x - (√3/2)y

and

y' = (√3)x' + 2y replace x' by (1/2)x - (√3/2)y

y' = (√3)[(1/2)x - (√3/2)y] + 2y

y' = (√3/2)x - (3/2)y + 2y

y' = (√3/2)x + (1/2)y

Thus, the rotated point (if the angle of rotation about the origin is 60 degrees clockwise) is [(1/2)x - (√3/2)y, (√3/2)x + (1/2)y].