解:由:a//b,得:sin⊙/cos⊙ = 2/1,即:tan⊙ = 2.又知:其中⊙∈（0,派/2).
故:cos⊙=1/根号[1+(tan⊙)^2]=1/根号5 .
sin⊙=tan⊙*[cos⊙]=2*[1/根号5]= 2/根号5.下一个问呢？？cosw = cos[⊙+(w-⊙)] = cos⊙*cos(w-⊙) - sin⊙ *sin(w-⊙)= [1/根号5 ]*(4/5)-[2/根号5 ]*(3/5)= (4-6)/[5*根号5] =- 2/5根号5.sin（Θ-w）=3/5，和sin(w-⊙)= 3/5一样吗？？是我没看清楚.若:sin（Θ-w）=3/5，则sin(w-⊙)= -3/5.故,应当更正为:cosw = cos[⊙-(⊙-W)] = cos⊙*cos(⊙-w) + sin⊙ *sin(⊙-w)= [1/根号5 ]*(4/5)+[2/根号5 ]*(3/5)= (4+6)/[5*根号5] = 2/5根号5.I am very sorry!(4+6)/[5*根号5] = 2/5根号5.应改为(4+6)/[5*根号5] = 2/根号5你说的正确!