求值:[2sin50°+sin10°(1+√3tan10°)]√(1-sin^2 10°)

1个回答

  • 1-sin²10=cos²10

    所以原式=(2sin50+sin10+√3sin10*sin10/cos10)*cos10

    =2sin50cos10+sin10cos10+√3sin²10

    =2sin50cos10+1/2*sin20+√3(1-cos20)/2

    =2sin50cos10+1/2*sin20-√3/2cos20+√3/2

    =2sin50cos10+sin20cos60-cos20sin60+√3/2

    =2sin50cos10+sin(20-60)+√3/2

    =2sin50cos10-sin40+√3/2

    =2sin50cos10-sin(50-10)+√3/2

    =2sin50cos10-sin50cos10+cos50sin10+√3/2

    =sin50cos10+cos50sin10+√3/2

    =sin(50+10)+√3/2

    =sin60+√3/2

    =√3