f(x)=2sin(π/2-x/2)sin(π+x/2)+cos方(π/2-x/2)-cos方(π+x/2)
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f(x)=2sin(π/2-x/2)sin(π+x/2)+cos方(π/2-x/2)-cos方(π+x/2)
1、若x属于(0,π/2),求f(x)的最小值
2、设g(x)=f(2x-π/4)+2m,x属于【π/4,7π/8】,若g(x)有两个零点,求实数m的取值范围.
1、若x属于(0,π/2),求f(x)的最小值
2、设g(x)=f(2x-π/4)+2m,x属于【π/4,7π/8】,若g(x)有两个零点,求实数m的取值范围.
解析:
f(x)=2sin(π/2-x/2)sin(π+x/2)+cos²(π/2-x/2)-cos²(π+x/2)
=2cos(x/2)*[-sin(x/2)] +sin²(x/2)-cos²(x/2)
=-sinx -cosx
= - (根号2)*sin(x+π/4)
(1).若x属于(0,π/2),那么:π/4
f(x)=2sin(π/2-x/2)sin(π+x/2)+cos²(π/2-x/2)-cos²(π+x/2)
=2cos(x/2)*[-sin(x/2)] +sin²(x/2)-cos²(x/2)
=-sinx -cosx
= - (根号2)*sin(x+π/4)
(1).若x属于(0,π/2),那么:π/4
f(x)=2sin(π/2-x/2)sin(π+x/2)+cos方(π/2-x/2)-cos方(π+x/2)
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