1.化简:sin80°(1+√3tan10°)2.若tan(α+π/4)=-3/5,则tan(α-π/4)=?
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1.化简:sin80°(1+√3tan10°)
2.若tan(α+π/4)=-3/5,则tan(α-π/4)=?
(1)sin80°(1+√3tan10°)
=cos10°(1+√3sin10°/cos10°)
=cos10°[(cos10°+√3sin10°)/cos10°]
=2(1/2cos10°+√3/2sin10°)
=2(sin30°cos10°+cos30°sin10°)
=2sin40°
本题原题怀疑是sin50°而不是sin80°
若是sin50°则答案是1
sin50°(1+√3tan10°)
=sin50°(1+√3sin10°/cos10°)
=sin50°[(cos10°+√3sin10°)/cos10°]
=2sin50°(1/2cos10°+√3/2sin10°)/cos10°
=2sin50°(sin30°cos10°+cos30°sin10°)/cos10°
=2sin50°sin40°/cos10°
=2cos40°sin40°/cos10°
=sin80°/cos10°
=cos10°/cos10°
=1
(2)tan(α-π/4)
=tan[π/2-(α+π/4)]
=sin[π/2-(α+π/4)]/cos[π/2-(α+π/4)]
=cos(α+π/4)/sin(α+π/4)
=1/tan(α+π/4)
=-5/3
1.化简:sin80°(1+√3tan10°)
=sin80°+√3cos80°
=2(1/2sin80°+√3/2cos80°
=2sin(80+60)
=2sin40
2
tan(α+π/4)=-3/5,
(tana+1)/(1-tana)=-3/5则
tan(α-π/4)=(tana-1)/(1+tana)=-5/3
化简√1-sin80°=?
1/sin10°-√3/sin80°
求值:1/sin10°-√3/sin80°
1/sin10°— √3/sin80°
求值[2sin50°+sin10°(1+√3tan10°)]*√2(sin80°)*(sin80°)
[2sin50°+sin10°(1+√3tan10°)]*√2(sin80°)*(sin80度)
化简【2sin50°+sin80°(1+√3tan10°)】/cos5°
计算(1/cos80)-(√3/sin80)
(数学)化简:[2sin50°+sin80°(1+√3tan10)]/√(1+cos10°)
1.化简:sin80°(1+√3tan10°)2.若tan(α+π/4)=-3/5,则tan(α-π/4)=?
化简(1-sin80)/cos80
求证:3+tan1°tan2°+tan2°tan3°=tan3°/tan1详细过程1°
tan1等于tan1°吗
求值:[2sin50°+sin10°(1+√3tan10°)]×√2sin80°
求值:[2sin50°+sin80°(1+√3tan10°)]/√2cos5°
求[2sin50°+sin10°(1+√3tan10°0]sin80°的值
[sin50°(cos10°+√3sina10°)]/sin80°
1/sin10°-√3/sin80°=______________我用计算机算出来是4