已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π
问题描述：
已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π（2）（sin2x-2cos^2x）/（1-tanx）
答
是sin(x+π/3)cos(x-π/3)+cos(x+π/3)sin(x-π/3)=-(2√2/3)吧?
如果是,则上式可化为sin2x=-2√2/3,
∵π(1) tanx=sin2x/(1+cos2x)=-√2.
(2) 2cos²x=1+cos2x=2/3,
∴原式=(-2√2/3-2/3)/(1-√2)=(6+4√2)/3.