高一数学---关于三角恒等变换---难

1个回答

  • 1、原式=(sin5/cos5-cos5/sin5)cos70/(1+sin70)

    =-(cos10/sin5cos5)cos70/(1+sin70)

    =-2cos10cos70/[sin10(1+sin70)]

    =-(cos80+cos60)/(sin10+sin70sin10)

    =-(sin10+1/2)/[sin10-(1/2)cos80-(1/2)cos60]

    =-(sin10+1/2)/[(1/2)sin10+1/4]

    =-(sin10+1/2)/[(1/2)(sin10+1/2)]

    =-2

    2、利用根与系数关系

    f(x)=1+sin(x/2)+(3^1/2-1)sin(x/2)-(1+(cos(x/2))

    =3^(1/2)*sin(x/2)-cos(x/2)

    =2[sin(x/2)cos30-cos(x/2)sin30]

    =2sin(x/2-30度)