Hello, Sconts!

I can help you with #1 . . .

1. Show that the coordinates of point (x,y) rotated counter-clockwise by an angle of θ

around the origin have the new coordinates (x·cosθ - y·sinθ, x·sinθ + y·cosθ).

Draw a picture that clearly illustrates your derivation, and provide plenty of explanation.

Hint: consider the distance from the origin to (x,y) as "d", and use the angle-sum formulas. Code:

| P'
| *(x',y')
| * :
| * :
| * :
| d * : P
| * : *(x,y)
| * d * :
| * θ * : :
| * * φ : :
- * - - - - - - - - + - - + - X
O

Point P(x,y) make angle φ with the x-axis.

Point P'(x',y') makes angle P'OP = θ.

Let d = OP = OP'.

The coordinates of P are: .x = d·cosφ, .y = d·sinφ .**[1]**

The coordinates of P' are: .x' = d·cos(θ + φ), .y' = d·sin(θ + φ)

We have: .x' .= .d·cos(θ + φ) .= .d·cosφ·cosθ - d·sinφ·sinθ

. . . . . . . . . . . . . . . . . . . . . . . . .↓ . . . - . - . . .↓

Substitute [1]: . . . . . . . . .x' .= . .x · cosθ . .- . .y · sinθ

. . Therefore: .**x' .= .x·cosθ - y·sinθ**

We have: .y' .= .d·sin(θ + φ) .= .d·cosφ·sinθ + d·sinφ·cosθ

. . . . . . . . . . . . . . . . . . . . . . . . .↓ . - . - . - . - .↓

Substitute [1]: . . . . . . . . .y' .= . .x · sinθ . .+ . .y · cosθ

. . Therefore: .**y' .= .x·sinθ + y·cosθ**