x属于(0,π/2),求[(sinx)2+1/(sinX)2][(cosX)2+1/(cosX)2]的最小值
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![x属于(0,π/2),求[(sinx)2+1/(sinX)2][(cosX)2+1/(cosX)2]的最小值](/uploads/image/z/5347928-56-8.jpg?t=x%E5%B1%9E%E4%BA%8E%EF%BC%880%2C%CF%80%2F2%29%2C%E6%B1%82%5B%28sinx%292%2B1%2F%28sinX%292%5D%5B%28cosX%292%2B1%2F%28cosX%292%5D%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC)
x属于(0,π/2),求[(sinx)2+1/(sinX)2][(cosX)2+1/(cosX)2]的最小值
x属于(0,π/2),求[(sinx)2+1/(sinX)2][(cosX)2+1/(cosX)2]的最小值
x属于(0,π/2),求[(sinx)2+1/(sinX)2][(cosX)2+1/(cosX)2]的最小值
[[(sinx)2+1/(sinX)2][(cosX)2+1/(cosX)]
=sin²xcos²x +cos²x/sin²x +sin²x/cos²x +1/(sin²xcos²x)
=sin²xcos²x + [(cos²x)²+(sin²x)²]/(sin²xcos²x) +1/(sin²xcos²x)
=sin²xcos²x +(sin²x+cos²x)²/(sin²xcos²x) -2 +1/(sin²xcos²x)
=sin²xcos²x+2/(sin²xcos²x) -2
令t=sin²xcos²x=1/4(sin2x)²
那么t∈(0,1/4]
原式子=t+2/t -2
设f(t)=t+2/t -2
f'(t)=1-2/t²=(t²-2)/t²,当t∈(0,1/4],f'(t)