因为：sin50°=sin(60°-10°)=sin60°cos10°-cos60°sin10°=(√3/2)cos10°-(1/2)sin10°所以，sin^50°=[(√3/2)cos10°-(1/2)sin10°]^=(3/4)cos^10°+(1/4)sin^10°-(√3/2)sin10°cos10°………………（1）同理，cos50°=cos(60°-10°)=cos60°cos10°+sin60°sin10°=(1/2)cos10°+(√3/2)sin10°所以，cos^50°=[(1/2)cos10°+(√3/2)sin10°]^=(1/4)cos^10°+(3/4)sin^10°+(√3/2)sin10°cos10°………………（2）将(1)(2)代入到原式中，得到：cos^10°-sin^50°+cos^50°-sin^10°=cos^10°-(3/4)cos^10°-(1/4)sin^10°+(√3/2)sin10°cos10°+(1/4)cos^10°+(3/4)sin^10°+(√3/2)sin10°cos10°-sin^10°=(1/2)cos^10°-(1/2)sin^10°+√3sin10°cos10°=(1/2)(cos^10°-sin^10°)+(√3/2)(2sin10°cos10°)=(1/2)cos20°+(√3/2)sin20°=sin30°cos20°+cos30°sin20°=sin50°

解:cos²10°-sin²50°+cos²50°-sin²10°=(cos²10°-sin²10°)+(cos²50°-sin²50°)=cos20°+cos100°=cos(60°-40°)+cos(60°+40°)=cos60°cos40°+sin60°sin40°+cos60°cos40°-sin60°sin40°=2cos60°cos40°=cos40°