解:由:a//b,得:sin⊙/cos⊙ = 2/1,即:tan⊙ = 2.又知:其中⊙∈（0,派/2).
故:cos⊙=1/根號[1+(tan⊙)^2]=1/根號5 .
sin⊙=tan⊙*[cos⊙]=2*[1/根號5]= 2/根號5.下一個問呢？？cosw = cos[⊙+(w-⊙)] = cos⊙*cos(w-⊙) - sin⊙ *sin(w-⊙)= [1/根號5 ]*(4/5)-[2/根號5 ]*(3/5)= (4-6)/[5*根號5] =- 2/5根號5.sin（Θ-w）=3/5，和sin(w-⊙)= 3/5一樣嗎？？是我沒看清楚.若:sin（Θ-w）=3/5，則sin(w-⊙)= -3/5.故,應當更正為:cosw = cos[⊙-(⊙-W)] = cos⊙*cos(⊙-w) + sin⊙ *sin(⊙-w)= [1/根號5 ]*(4/5)+[2/根號5 ]*(3/5)= (4+6)/[5*根號5] = 2/5根號5.I am very sorry!(4+6)/[5*根號5] = 2/5根號5.應改為(4+6)/[5*根號5] = 2/根號5你說的正確!