Question

已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x 求f(π/12)的值？

Answer 1

f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos²x
=根号下3sin2x/2+cos2x/2-cos2x/2+根号下3sin2x/2+2cos²x-1+1
=根号下3sin2x+cos2x+1
=2(根号下3sin2x/2+cos2x/2)+1
=2sin(2x+π/6)+1
f(π/12)=2sin(π/3)+1=根号下3+1,2,f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x
f(π/12)=sin(π/6+π/6)-cos(π/6+π/3)+2cos^2π/12-1+1
=sinπ/3-cosπ/2+cosπ/6+1
=1/2根号3+1/2根号3+1
=根号3+1,2,f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x ；
f(π/12)=sin(π/6+π/6)-cos(π/6+π/3)+2cos^2(π/12)-1+1;
=sin(π/3)-cos(π/2)+cos(π/6)+1
=√3/2-0+√3/2+1
=√3+1,1,f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos^2x
f(π/12)=sin(π/3)-cos(π/2)+ (1+cos(π/6))
=根号(3)/2 - 0 + 1 + 根号(3)/2
=根号(3) + 1,0,