x∈[π/4,π/3],f(x)=1/2sin(x-5π/12)cos(5π/12-x)+[(根号3)/2][sin(x

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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

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

x∈[π/4,π/3],f(x)=1/2sin(x-5π/12)cos(5π/12-x)+[(根号3)/2][sin(x 已知函数f(x)=sin^2*x-根号3*sinπ/4*x*cosπ/4*x 已知函数f(x)=sin²ωx+根号3sinωx乘sin(ωx+π/2)+2cos²ωx,x∈R,（求化简f(x)=sin^2x+2根号3sin(x+π/4)cos(x-π/4)-cos^2x-根号3 f(x)=sin平方x+2根号3sin(x+π/4)cos(x-π/4)-cos平方x-根号3 化简2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1 已知函数f(x)=根号3sin(2x-π/6)+2sin的平方（x-π/12)(x∈R)(1)求函数f(x) 已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x 已知函数f(x)=cos(2x-π\3)+sin²x-cos²x f(x)=2sin^2(π/4-x)-2根号3cos^2x+根号3如题 f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx 已知函数f(x)=根号3sinπx+cosπx,x属于R