问题描述：
x∈[π/4,π/3],f(x)=1/2sin(x-5π/12)cos(5π/12-x)+[(根号3)/2][sin(x+4)]^2
化简一下就行
问题解答：
我来补答
I suppose the last terms is [sin(x+π/4)]^2
f(x)
=1/2sin(x-5π/12)cos(5π/12-x)+[√3/2][sin(x+π/4)]^2
= (-1/4)(2sin(5π/12-x)cos(5π/12-x) +[√3/2][sin(x+π/4)]^2
=(-1/4)(sin2(5π/12-x))+[√3/2][sin(x+π/4)]^2
=(-1/4)(sin(π-(π/6+2x)) +[√3/2][sin(x+π/4)]^2
=(-1/4)sin(π/6+2x)+ [√3/2](1-cos2(x+π/4))/2)
=(-1/4)sin(π/6+2x)+ √3/4- (√3/4)cos(π/2+2x)
=(-1/4)sin(π/6+2x)+ √3/4+ (√3/4)sin2x
=(-1/4)( sin(π/6)cos2x+cos(π/6)sin2x) +√3/4+ (√3/4)sin2x
=(-1/4)( (1/2)cos2x+(√3/2)sin2x) +√3/4+ (√3/4)sin2x
=(-1/8)cos2x + (√3/8)sin2x+ √3/4
=1/4(cos(π/6)sin2x - sin(π/6)cos2x) + √3/4
= (1/4)(sin(2x-π/6)) +√3/4
