求值:[sin50°(1+tan10°√3)-cos20°]/[cos80°乘以根号下1-cos20°]
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/05/09 19:25:04
求值:[sin50°(1+tan10°√3)-cos20°]/[cos80°乘以根号下1-cos20°]
(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°)-cos20°)/(cos80°*√(2(sin10°)^2))
=(sin50°(sin30°*cos10°+cos30°*sin10°)/(1/2)*cos10°-cos20°)/((√2)cos80°*sin10°)
=(sin50°*sin40°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=(sin50°*cos50°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)sin100°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)sin80°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)cos10/(1/2)*cos10°-cos20°)/((√2)sin10°sin10°)
=(1-cos20°)/((√2)sin10°sin10°)
=2(sin20°)^2/(√2)(sin20°)^2
=2/√2
=√2
=(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°)-cos20°)/(cos80°*√(2(sin10°)^2))
=(sin50°(sin30°*cos10°+cos30°*sin10°)/(1/2)*cos10°-cos20°)/((√2)cos80°*sin10°)
=(sin50°*sin40°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=(sin50°*cos50°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)sin100°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)sin80°/(1/2)*cos10°-cos20°)/((√2)sin10°*sin10°)
=((1/2)cos10/(1/2)*cos10°-cos20°)/((√2)sin10°sin10°)
=(1-cos20°)/((√2)sin10°sin10°)
=2(sin20°)^2/(√2)(sin20°)^2
=2/√2
=√2
求值:[sin50°(1+tan10°√3)-cos20°]/[cos80°乘以根号下1-cos20°]
求值:【sin50°[1+根号(3)tan10°]—cos20°】/[cos80°根号(1—cos20°)】
化简:sin50º(1+√3tan10°)-cos20°/cos80°√(1-cos20°)
化简:【sin50°·(1+√3tan10° )-cos20°】/(cos80°· √1-cos20°)
化简sin50°(1+√3tan10°)-cos20°/cos80°√(1-cos20°)
(sin50(1+根号3*tan10)-cos20)/cos80根号(1-cos20)
【跪求】【求值】(sin50(1+根号(3)*tan10)-cos20)\(根号(2)*cos80*sin10)
sin50度(1+根号3tan10度)-cos20度/cos80度根号1-cos20度=
cos80度根号1-cos20度分之sin50度(1+根号3tan10度)-cos20度=
化简{2sin50°+sin10°【1+(根号3tan10°)】根号(1加cos20)°
sin50(1+根号3倍的tan10)-cos20,再整体除以cos80*根号下(1-cos20)求这个值是多少,关于三
化简[2sin50°+sin10°(1+3tan10°)]1+cos20°