急!已知sinα=1/3,则sin2（π/4-α）=?

7\9我有个问题···为什么cosα一定是正的····sin2（π/4-α）=sin｛2（π/4-/α）｝ =sin｛2π/4-2α｝ =sin｛π/2-2α｝ =sin(π/2)cos(2α)-cos(π/2)sin(2α)=1cos(2α)-0sin(2α)=cos(2α)-0=cos(2α)=cosα*cosα-sinα*sinα=1-(sinα)^2--(sinα)^2=1-2(sinα)^2=1-2(1\3)^2=1-2X1\9=1-2\9=7\9