已知1－tana分之1+tana=3+2√2,求cos（π－a）+sin（π+a）cos（π－a）+2sin(a－π)的值.

首先化简式子原式=(-cos(a))+sin(a)×(-cos(a))+2(-sin(a)) =cos(a)-sin(a)cos(a)+2sin(a) =(cos(a)-sin(a)cos(a)+2sin(a))/1 =(cos(a)-sin(a)cos(a)+2sin(a))/(sin(a)+cos(a)) 然后等式两边同时除以cos(a) =(2tan(a)-tan(a)+1)/(tan(a)+1) 因为(1+tan(a))/(1-tan(a))=3+2√2 ===>tan(a)=√2/2 代入计算可得 (2×1/2-√2/2+1)/(1/2+1)=(4-√2)/3