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[2sin130º+sin100º(1+根号3tAn370&Am...

2sin30°+sin100°(1+根号3倍的tan370°) =1+sin80°(1+根号3倍的tan10°) =1+cos10°(cos10°+根号3倍的sin10°)/cos10° =1+(cos10°+根号3*sin10°) =1+2(1/2*cos10+根号3/2*的sin10) =1+2(sin30°*cos10°+cos30°sin10°) =1+2sin40°

答案:2 解: [2sin130+sin100(1+根号3tan370)]/根号(1+cos10) = [2sin50+cos10(1+根号3tan10)]/根号(1+cos10) = [2sin50+cos10+根号3sin10)]/根号(1+cos10) =[2sin50+2{[(1/2)]*cos10+[(根号3)/2]*sin10)]/根号(1+cos10) =2(sin50+cos50)/根号(1+cos10)=2*(根号2)*sin(50+45°)/根号(2cos5) =2*(根号2)*cos5°/[(根号2)*cos5]=2

(sin50°(1 √3*tan10°)-cos20°)/(cos80°*√(1-cos20°)) =(sin50°(1 √3*sin10°/cos10°)-cos20°)/(cos80°*√(1-cos20°)) =(sin50°((cos10° √3*sin10°)/cos10°)-cos20°)/(cos80°*√(2(sin10°)^2)) =(sin50°((1/2)cos10° (√3/2)*sin10°)/((1/2)*cos10°)-

原式=2sin50°+sin80°(1+ 3 tan10°) 2 cos5° =2sin50°+cos10°(1+ 3 sin10° cos10° ) 2 cos5° =2sin50°+cos10°+ 3 sin10° 2 cos5° =2sin50°+2(1 2 cos10°+ 3 2 sin10°) 2 cos5° =2sin50°+2sin(10°+30°) 2 cos5° =2(sin50°+sin40°) 2 cos5° =4sin45°cos5° 2 cos5° =4sin45° 2 =2 2 2 =2故答案为2.

sin130°(1+根号3tan190°) =sin50(1+√3tan10) =sin50(cos10+√3sin10)/cos10 =sin50*2(1/2cos10+√3/2sin10)/cos10 =sin50*2sin(30+10)/cos10 =2cos40*sin40/sin80 =sin80/sin80 =1

1+√3tan370=2(1/2+√3/2tan10)=2(sin30+cos30(sin10/cos10))=2sin40/cos10故原式=2cos40+cos10(2sin40/cos10)=2(sin40+cos40)

〔2*sin130°+sin100°(1+√3*tan370°)〕/√(1+cos10°) =(2sin50°+cos10°(1+√3tan10°))/√(1+cos10°) =(2sin50°+cos10°+√3sin10°)/√(1+cos10°)=2(sin50°+sin40°)/√(1+cos10°)=2(sin50°+cos50°)/√(1+cos10°)将原式平方=4(1+2sin50°cos50°)/(1+cos10°)=4(1+cos10°)/(1+cos10°)=4再求其算数平方根=2

[2sin130°+sin100°(1+√3*tan190°)]/√(1+cos10°)=[2sin50°+sin80°(1+√3*tan10°)]/√(1+cos10°)=[2sin50°+cos10°(1+√3*(sin10°/cos10°))]/√(1+cos10°)=[2sin50°+(cos10°+√3*sin10°)]/√(1+cos10°)=[2sin50°+(cos10°+√3*sin10°)]/√(1+

1+√3tan370°=1+√3tan10°sin130°=sin50° sin100°=sin80°=cos10°∴2sin130°+sin100°(1+√3tan370°)=2sin50°+cos10°(1+√3tan10°)=2sin50°+(cos10°+√3sin10°)=2sin50°+2sin(10°+30°)=2

-π/20, 而tanα= -1/3, 即1/(cosα)^2=1+(tanα)^2=10/9, 而cosα>0, 故cosα= 3/√10,而sinα= -1/√10 所以2sin^2α= 1/5, sin2α=2sinα*cosα= -3/5 cos(α-π/4)=cosα*cos(π/4) +sinα*sin(π/4)=1/√5 于是 2sin^2α+sin2α/cos(α-π/4) =1/5 - (3/5) / (1/√5

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