已知点P(x,y)绕原点旋转θ角到点P(x',y')如题,求证:x'=xcosθ-ysinθy'=xsinθ+ycosθ

Question

已知点P(x,y)绕原点旋转θ角到点P(x',y')
如题,
求证:x'=xcosθ-ysinθ
y'=xsinθ+ycosθ

Answer 1

设P点到原点的距离为r,OP与x轴正向所成的角为a
则有x=rcosa,y=rsina
旋转θ角,相当于r不变,把角a加上了θ角
则有x'=rcos(a+θ)=rcosa*cosθ-rsina*sinθ=xcosθ-ysinθ
,y=rsin(a+θ)=rsina*cosθ+rcosa*sinθ=xsinθ+ycosθ