【数学题】求函数y=sinx⼀2(sinx⼀2-cosx⼀2)的最大值和最小值

2025-03-07 08:07:20
推荐回答(1个)
回答1:

y=sin²(x/2)-sin(x/2)cos(x/2)
=(1-cosx)/2-(1/2)sinx
=1-(1/2)(sinx+cosx)
=1-(√2/2)sin(x+π/4)
当sin(x+π/4)=-1时,y有最大值,y(max)=1+√2/2;
当sin(x+π/4)=1时,y有最小值,y(min)=1-√2/2;

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