题目

高一数学---关于三角恒等变换---难
1.求值:(tan5度-tan85度)*cos70度/(1+sin70度) -2,
2.已知sin(x/4),cos(x/4)是y的方程y^2+py+q=0 的两个实根,设函数f(x)=p^2+2(3^1/2-1)q-2(cos(x/4))^2,如何化得f(x)=2sin(x/2-30度)

提问时间：2021-03-18

答案

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)