根据题目所给角度范围可得：cos(α-β)=-4/5 cos(α+β)=4/5 sin2α=sin[(α-β)+(α+β)]=sin(α-β)cos(α+β)+cos(α-β)sin(α+β)=24/25， cos2α=cos[(α-β)+(α+β)]=cos(α-β)cos(α+β)-sin(α-β)sin(α+β)=-1
α-β∈(π/2,π),α+β∈∈(3π/2,2π),→ sin(α-β)=3/5,→cos(α-β)=-4/5 sin(α+β)=-3/5,→cos(α+β)=4/5 2α∈(2π,3π) sin2α=sin[(α-β)+(α+β)]= sin(α-β)cos(α+β)+cos(α-β)sin(α+β)= (3/5)*(4/5)+(-4/5)*(-3/5)=24/25 cos2α=cos(α+β)cos(α-β)-sin(α+β)sin(α-β)== (-4/5)*(4/5)-(3/5)*(-3/5)=-7/25
答：=2cos60sin(x+60)+2sin(x-60)+2*sin60cos(x+60) =2sin(x+60+60)+2sin(x-60) =0详情>>
答：详情>>
