讲一下大概思路也行,泪流满面……没时间了……1.tan70°cos10°(1-√tan20°)2.(tan10°-√3)•(cos10°/sin50°) 3.3sin²[（A+B）/2 ]+ cos² [（A-B）/2]=2,且cosA•cosB ≠0,求tanAtanB的值 4.化简cos²(x/2-7/8π)- cos² (x/2+7/8π)

第一题是不是题目错了应该是
1.tan70°cos10°(1-√3*tan20°)
=tan70*cos10(1-√3*sin20/cos20)
=2tan70*cos10[1/2-(√3/2)sin20/cos20]
=2tan70*cos10[(1/2)cos20-(√3/2)sin20]/cos20
=2tan70*cos10[sin30cos20-cos30sin20]/cos20
=2tan70*cos10*sin(30-20)/cos20=2cot20*cos10*sin10/cos20
=cot20*sin20/cos20=1
2.(tan10°-√3)?(cos10°/sin50°)
=2(sin10/cos10-√3)(cos10/sin50)
=(2/cos10)(1/2sin10-√3/2cos10)(cos10/sin50)
=(2/cos10)*sin(10-60)*(cos10/sin50)
=(-2sin50/cos10)*(cos10/sin50)
=-2
3.
3sin?[（A+B）/2 ]+ cos?[（A-B）/2]
=(-3/2)*(1-2sin?[（A+B）/2 ])+
3/2+1/2*(2cos?[（A-B）/2]-1)+1/2
=(-3/2)*cos(A+B)+1/2*cos(A-B)+2=(-3/2)*(cosAcosB-sinAsinB)+1/2*(cosAcosB+sinAsinB)+2
=-cosAcosB+2sinAsinB+2=2
所以有
cosAcosB=2sinAsinB
tanAtanB=1/2
4.
cos?(x/2-7/8π)- cos?(x/2+7/8π)
=1/2*[2cos?(x/2-7/8π)-1]+1/2-
1/2*[2 cos?(x/2+7/8π)-1]-1/2
=1/2*[cos(x-7/4π)-cos(x+7/4π)]
=sinx*sin7/4π=-√2/2*sinx
终于完成了!
新年快乐!