[cos40+sin50(1+√3tan10)]/sin70√(1+sin50) 希望有具体过程 .

1个回答

  • 先计算sin50(1+√3tan10)的值:

    sin50(1+√3tan10)

    =sin50[1+√3(sin10/cos10)]

    =sin50[(cos10+√3sin10)/cos10]

    =sin50[2sin(30+10)/cos10]

    =(2sin50sin40)/cos10°

    =(2cos40°*sin40°)/cos10°

    =sin80°/cos10°

    =cos10°/cos10°

    =1

    所以[cos40+sin50(1+√3tan10)]/sin70√(1+sin50)

    =[cos40+1]/[sin70√(1+sin50)]

    =(2 cos²20)/[ cos20*√(sin25+cos25)²]

    =(2 cos²20)/[ cos20*(sin25+cos25)]

    =(2 cos20)/(sin25+cos25)

    =(2 cos20)/[√2(sin25+45))

    =(2 cos20)/[√2sin70]

    =(2 cos20)/[√2cos20]

    =√2.