If $x=a$·cosθ + b·sinθ and $y=a$·sinθ − b·cosθ, then $x^2+y^2$ is equal to:

Square each and add: $x^2=a^2$co$s^2$θ + $2ab$·sinθ·cosθ + $b^2$si$n^2$θ and $y^2=a^2$si$n^2$θ − $2ab$·sinθ·cosθ + $b^2$co$s^2$θ. The cross terms are equal and opposite, so they vanish, leaving $a^2$(co$s^2$θ + si$n^2$θ) + $b^2$(si$n^2$θ + co$s^2$θ) = $a^2+b^2$ — a constant, so θ has been eliminated. Check with $a=3,b=4$, θ = 37°: $x=2.396+2.407=4.803$ and $y=1.806-3.195=-1.390$, and 23.07 + 1.93 = 25.00 = 9 + 16. Missing the cancellation leaves the θ-dependent $a^2+b^2+2ab$·sinθ·cosθ, keeping only the cross terms gives $2ab$, and subtracting instead of adding the two squares gives $a^2-b^2$.