解cos2(π-α)+sin(π+α)cos(π-α)+2sin2(α-π)=cos（2π-2α)+[-sin(α)][-cos(α)]+2sin(2α-2π)=cos2α+sin(α)cos(α)-2sin(2π-2α)=cos2α+sin(α)cos(α)-2sin(-2α)=cos2α+1/2*2*sin(α)cos(α)+2sin(2...我的題目是cos*2（π-α）+sin（π+α）cos（π-α）+2【sin*2（α-π）】。我表述不清...麻煩你再做一遍~~~cos*2（π-α）+sin（π+α）cos（π-α）+2【sin*2（α-π）】=cos²（π-α）+sin（π+α）cos（π-α）+2sin²（α-π）=[-cosa]²+[-sin(α)][-cos(α)]+2sin²（π-a）=cos²a+sin(α)cos(α)+2sin²a=cos²a+1/2*2sin(α)cos(α)+2sin²a=(cos²a+sin²a)+1/2*2sin(α)cos(α)+sin²a=1+1/2sin(2α)+sin²a=1+1/2*2tana/(1+tan²a)+sin²a=1+tana/(1+tan²a)+sin²a=1+tana/(1+tan²a)+1/(1+cot²a）=1+tana/(1+tan²a)+1/(1+（1/tana）²）=1+√2/2/(1+(√2/2)²)+1/(1+（1/√2/2）²）=1+√2/2/3/2+1/(1+2）=1+√2/3+1/3=(√2+4)/3