作业帮求函数F(x)=[in²X+1/(2010sin²X)][os²x+1/(2010cos²X)]的最小值
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![求函数F(x)=[in²X+1/(2010sin²X)][os²x+1/(2010cos²X)]的最小值](/uploads/image/z/5242861-37-1.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0F%28x%29%3D%5Bin%26%23178%3BX%2B1%2F%282010sin%26%23178%3BX%29%5D%5Bos%26%23178%3Bx%2B1%2F%282010cos%26%23178%3BX%29%5D%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC)
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求函数F(x)=[in²X+1/(2010sin²X)][os²x+1/(2010cos²X)]的最小值
求函数F(x)=[in²X+1/(2010sin²X)][os²x+1/(2010cos²X)]的最小值
求函数F(x)=[in²X+1/(2010sin²X)][os²x+1/(2010cos²X)]的最小值
令a=(sinx)^2,b=(cosx)^2
则F(x)=(a+1/2010a)(b+1/2010b)
=ab+a/2010b+b/2010a+1/(2010^2*ab)
=ab+[(a^2+b^2)/2010ab]+1/(2010^2*ab)
=ab+[(a+b)^2-2ab]/2010ab+1/(2010^2*ab)
=ab+(1-2ab)/2010ab+1/(2010^2*ab)
=ab+1/2010ab-1/1005+1/(2010^2*ab)
=ab+(2011/2010^2ab)-1/1005
>=2根号(2011/2010^2)-1/1005
=2根号[（2011）/2010] -1/1005
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