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  • (sin50(1+根號(hào)3*tan10)-cos20)/cos80根號(hào)(1-cos20)

    (sin50(1+根號(hào)3*tan10)-cos20)/cos80根號(hào)(1-cos20)
    數(shù)學(xué)人氣:556 ℃時(shí)間:2020-01-27 21:49:50
    優(yōu)質(zhì)解答
    (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
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