问题: 高三复习
解答:
f(x)=sin²x +sinxcosx
=(1-cos2x)/2 +(1/2)sin2x
=(1/2) +(1/2)(sin2x -cos2x)
=(1/2)+(√2/2)sin(2x -π/4)
x∈[0,π/2]
最小值 sin(2x -π/4)=-(√2)/2时
f(x)= 0
f(a)=1/2) +(1/2)(sin2a -cos2a) =3/4
==>
sin2a -cos2a =1/2 ........(1)
(sin2a -cos2a)²=1/4
2sin2acos2a =3/4>0 ==> 2a∈[0,π/2]
(sin2a+cos2a)² =7/4
sin2a+cos2a =√7/2 .......(2)
(1),(2)==>sin2a =(2+√7)/8
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