已知函数y=1/2cos平方x+√3/2sinxcosx+1,x属于R,求它的振幅周期初相.要详细过
1个回答

1/2=sin(π/6),√3/2=cos(π/6)

y=(1/2)(cosx)^2+(√3/2)sinxcosx+1

=cosx[sin(π/6)cosx+cos(π/6)sinx]+1

=sin(x+π/6)cosx+1 ………………………………………………………(1)

sin(2x+π/6)=sin(x+π/6+x)=sin(x+π/6)cosx+cos(x+π/6)sinx ………(2)

1/2=sin(π/6)=sin(x+π/6-x)=sin(x+π/6)cosx-cos(x+π/6)sinx ………(3)

(2)+(3)可得:sin(x+π/6)cosx=[sin(2x+π/6)]/2+1/4 ……………(4)

把(4)代入(1)继续化简:

sin(x+π/6)cosx+1

=(1/2)sin(2x+π/6)+1/4+1

=(1/2)sin(2x+π/6)+5/4

因此:y=(1/2)sin(2x+π/6)+5/4

所以振幅是1/2,周期是2π/2=π,初相是π/6