若sin(x+π\3)=1\4,sin(π\6+2x)的值

若sin(x+π\3)=1\4,sin(π\6+2x)的值
若sin(x+π\3)=1\4,sin(π\6+2x)的值

若sin(x+π\3)=1\4,sin(π\6+2x)的值
sin(x+π/3)=(1/2)sinx+(√3/2)cosx=1/4,2sinx+2√3cosx=1
对上式进行平方得：3+8√3sinxcosx+8(cosx)^2=0,即√3sinxcosx+(cosx)^2=-3/8
sin(2x+π/6)=sin2xcosπ/6+cos2xsinπ/6=(√3/2)sin2x+(1/2)cos2x
=(√3/2)2sinxcosx+(1/2)*[2(cosx)^2-1]=√3sinxcosx+(cosx)^2-1/2
=-3/8-1/2=-7/8