化简√1-sinx+√1+sinxx∈(0,π/2)
化简√1-sinx+√1+sinxx∈(0,π/2)
【正确答案】:原式化为√[sin(x/2)-cos(x/2]2+√[sin(x/2)+cos(x/2]2 =∣[sin(x/2)-cos(x/2]∣+∣[sin(x/2)+cos(x/2]∣ 因为x∈(0,π/2),所以x/2∈(0,π/4),0
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