Question

高中数学，三角函数..化简..急求~！！

化简:[根号(1-2sin290°cos110°)]/(sin250°+sin20°)

Answer 1

sin290= -sin70= -cos20，cos110= -sin20 又cos20>sin20 所以分子变为cos20-sin20
sin250= -sin110= -cos20，所以分母为sin20-cos20
所以结果为 -1
谢谢采纳！

Answer 2

[根号(1-2sin290°cos110°)]/(sin250°+sin20°)
=[根号(1-2sin(180°+110°)cos110°)]/[sin(270°-20°)+sin20°]
=[根号(1+2sin110°cos110°)]/(-cos20°+sin20°)
={根号[(sin110°)^2+(cos110°)^2+2sin110°cos110°]}/(-cos20°+sin20°)
={根号[(sin110°+cos110°)^2]}/(-cos20°+sin20°)
=(sin110°+cos110°)/(-cos20°+sin20°)
=(cos20°-sin20°)/(-cos20°+sin20°)=-1

Answer 3

原式=√(sin²110+cos²110+2sin110cos110)/(-sin70+cos70)
=√(sin110+cos110)²/(-sin70+cos70)
=√(sin70-cos70)²/(-sin70+cos70)
=|sin70-cos70|/(-sin70+cos70)
=(sin70-cos70)/(-sin70+cos70)
=-1

Answer 4

sin290°=sin(2π-70°)=-sin70°
cos110°=cos(π-70°)=-cos70°
sin250°=sin(π+70°)=-sin70°
sin20°=cos70°
故原式=√(1-2sin70°cos70°)/(cos70°-sin70°)=|sin70°-cos70°|/(cos70°-sin70°)=-1

Answer 5

根号(1-2sin290°cos110°)]/(sin250°+sin20°)=根号（1+2sin110ºcos110º)/(-sin70º+sin20º)
=根号(1-2sin20ºcos20º)/(sin20º-cos20º)=根号(sin²20º+cos²20º-2sin20ºcos20º)/(sin20º-cos20º)
=-1