作业帮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.化简：sin80°(1+√3tan10°)
2.若tan(α+π/4)=-3/5,则tan（α-π/4)=?
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