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
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![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](/uploads/image/z/2823254-62-4.jpg?t=%E8%AE%B2%E4%B8%80%E4%B8%8B%E5%A4%A7%E6%A6%82%E6%80%9D%E8%B7%AF%E4%B9%9F%E8%A1%8C%2C%E6%B3%AA%E6%B5%81%E6%BB%A1%E9%9D%A2%E2%80%A6%E2%80%A6%E6%B2%A1%E6%97%B6%E9%97%B4%E4%BA%86%E2%80%A6%E2%80%A61.tan70%C2%B0cos10%C2%B0%281-%E2%88%9Atan20%C2%B0%292.%28tan10%C2%B0-%E2%88%9A3%29%26%238226%3B%28cos10%C2%B0%2Fsin50%C2%B0%29+3.3sin%26sup2%3B%5B%EF%BC%88A%2BB%EF%BC%89%2F2+%5D%2B+cos%26sup2%3B+%5B%EF%BC%88A-B%EF%BC%89%2F2%5D%3D2%2C%E4%B8%94cosA%26%238226%3BcosB+%E2%89%A00%2C%E6%B1%82tanAtanB%E7%9A%84%E5%80%BC+4.%E5%8C%96%E7%AE%80cos%26sup2%3B%28x)
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
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)
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
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!