求值:(1+cos20°)/(2sin20°)-sin10°(tan^-1 5°-tan5°)(解题过程)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/28 03:17:06
求值:(1+cos20°)/(2sin20°)-sin10°(tan^-1 5°-tan5°)(解题过程)
sinθ+sinφ=2sin(θ/2+θ/2)cos(θ/2-φ/2)
sinθ-sinφ=2cos(θ/2+φ/2)sin(θ/2-φ/2)
cosθ+cosφ=2cos(θ/2+φ/2)cos(θ/2-φ/2)
cosθ-cosφ=-2sin(θ/2+φ/2)sin(θ/2-φ/2)
sinαsinβ=-1/2[cos(α+β)-cos(α-β)]
cosαcosβ= 1/2[cos(α+β)+cos(α-β)]
sinαcosβ= 1/2[sin(α+β)+sin(α-β)]
cosαsinβ= 1/2[sin(α+β)-sin(α-β)]
((1+cos20°)/2sin20°)-sin10°(cot5°-tan5°) =[(1+cos20°)/4sin10°cos10°]-sin10°(cot5°-tan5°)
= [(1+cos²10°-sin²10)/4sin10°cos10°]-sin10°(cot5°-tan5°)
=2cos²10°/4sin10°cos10°-sin10°(cot5°-tan5°)
=(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°)
=(cos10°/2sin10°)-2(cosp²5°-sin²5°)
=(cos10°/2sin10°)-2cos10°
=(cos10°-4sin10°cos10°)/2sin10°
=(sin80°-2sin20°)/2sin10°
=[(sin80°-sin20°)-sin20°]/2sin10°
=(2cos50°sin30°-sin20°)/2sin10°
=(sin40°-sin20°)/2sin10°
=2cos30°sin10°/2sin10°
=cos30°
=√3/2
sinθ-sinφ=2cos(θ/2+φ/2)sin(θ/2-φ/2)
cosθ+cosφ=2cos(θ/2+φ/2)cos(θ/2-φ/2)
cosθ-cosφ=-2sin(θ/2+φ/2)sin(θ/2-φ/2)
sinαsinβ=-1/2[cos(α+β)-cos(α-β)]
cosαcosβ= 1/2[cos(α+β)+cos(α-β)]
sinαcosβ= 1/2[sin(α+β)+sin(α-β)]
cosαsinβ= 1/2[sin(α+β)-sin(α-β)]
((1+cos20°)/2sin20°)-sin10°(cot5°-tan5°) =[(1+cos20°)/4sin10°cos10°]-sin10°(cot5°-tan5°)
= [(1+cos²10°-sin²10)/4sin10°cos10°]-sin10°(cot5°-tan5°)
=2cos²10°/4sin10°cos10°-sin10°(cot5°-tan5°)
=(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°)
=(cos10°/2sin10°)-2(cosp²5°-sin²5°)
=(cos10°/2sin10°)-2cos10°
=(cos10°-4sin10°cos10°)/2sin10°
=(sin80°-2sin20°)/2sin10°
=[(sin80°-sin20°)-sin20°]/2sin10°
=(2cos50°sin30°-sin20°)/2sin10°
=(sin40°-sin20°)/2sin10°
=2cos30°sin10°/2sin10°
=cos30°
=√3/2
求值:cos20°sin20°
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√(sin²70°)-√(1-2sin20°cos20°)+|sin20°-1|
sin40°*(1-sin10°)/[sin10°*(cos20°)^2]
1/cos20°-根号3/sin20°
化简((cos20°/sin20°)cos10°)+根号3(sin10°tan70°)-2cos40°
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化简根号下1-2sin20°cos20°