高中数学题,求解答!!
1.化简sin4x/[4sin^2(π/4+x)tan(π/4-x)]2.已知非零实属a,b,c满足c^2=a^2+b^2,又bn+am+2c=0,则m^2+n^2的最小...
1.化简sin4x/[4sin^2(π/4+x)tan(π/4-x)]
2.已知非零实属a,b,c满足c^2=a^2+b^2,又bn+am+2c=0,则m^2+n^2的最小值 展开
2.已知非零实属a,b,c满足c^2=a^2+b^2,又bn+am+2c=0,则m^2+n^2的最小值 展开
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sin4x/[4sin^2(π/4+x)tan(π/4-x)]
=sin4x/[4sin^2(π/4+x)tan[π/2-(π/4+x)]
=sin4x/[4sin^2(π/4+x)tan[π/2-(π/4+x)]
=sin4x/[4sin^2(π/4+x)cot(π/4+x)
=sin4x/[4sin^2(π/4+x)cos(π/4+x)/sin(π/4+x)
=sin4x/4sin(π/4+x)cos(π/4+x)
=sin4x/2sin(π/2+2x)c
=sin4x/2sin(π/2+2x)
=sin4x/2cos2x
=2sin2x2cos2x/2cos2x
=2sin2x
2)
=sin4x/[4sin^2(π/4+x)tan[π/2-(π/4+x)]
=sin4x/[4sin^2(π/4+x)tan[π/2-(π/4+x)]
=sin4x/[4sin^2(π/4+x)cot(π/4+x)
=sin4x/[4sin^2(π/4+x)cos(π/4+x)/sin(π/4+x)
=sin4x/4sin(π/4+x)cos(π/4+x)
=sin4x/2sin(π/2+2x)c
=sin4x/2sin(π/2+2x)
=sin4x/2cos2x
=2sin2x2cos2x/2cos2x
=2sin2x
2)
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求答案的同时自己也要去思索啊 现在的学生真的是!
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