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

问题描述:

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

(1-2sin290°cos110°)]/(sin250°+sin20°)
=(1-2sin70^cos70)(-sin70+sin20)
=(1-sin40)(sin20-sin70)
=-(1-sin40)*2*cos45*sin25
=1

[根号(1-2sin290°cos110°)]/(sin250°+sin20°)
=[√(1-2cos20°sin20°)]/(-cos20°+sin20°)
=(cos20°-sin20°)/(-cos20°+sin20°)
=-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!

既然他舅答了,我就打个酱油吧~~~

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

:[根号(1-2sin290°cos110°)]/(sin250°+sin20°)
=:[根号(1-2sin(270-20)*cos(90+20))]/(sin(270-20)+sin20°)
=:[根号(1-2cos20 * sin20°)]/cos20°+sin20°)
=:[根号(sin^20+cos^220-2sin290°cos110°)]/( - cos20°+sin20°)
=:[cos20-sin20]/( - sin20°+sin20°)
=1

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

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

[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

=(1+2cos20sin20)/(-sin70+sin20)
=(1+sin40)/(sin20-cos20)
=(1+sin40)/根号(2)sin25
=1/2

化简:[√(1-2sin290°cos110°)]/(sin250°+sin20°)
原式=√[1-2sin(360°-70°)cos(180°-70°)]/[sin(270°-20°)+sin20°]
=[√(1-2sin70°cos70°)]/[-cos20°+sin20°]
=(sin70°-cos70°)/(sin20°-cos20°)
=(cos20°-sin20°)/(sin20°-cos20°)=-1

={根号【1-2Sin70(-Cos70)】}/(Sin110+Sin20)
=【根号(1+2Sin70*Cos70)】/(Cos20+Sin20)
=(Sin70+Cos70)/(Cos20+Sin20)
=(Cos20-Sin20)/(Cos20+Sin20)