1、cos的4次方α-sin的4次方α=多少?2、tanα=2,则tan（π/4+α）等于做少?要详细步骤 本人真的不大明白

1.cos的4次方α-sin的4次方α=[COS^2(α)-SIN^2(α)][COS^2(α)+SIN^2(α)]
=COS^2(α)-SIN^2(α)=COS(2α)
2TAN(π/4+α)=[TAN(π/4)+TANα]/[1-TANπ/4*TANα]=-3

（1）.(cosa)^4-(sina)^4=[(cosa)^2-(sina)^2][(cosa)^2+(sina)^2]=[(cosa)^2-(sina)^2]*1=cos2a
(2).tan(p/4+a)=(1+tana)/(1-tana)=-3

1.利用两数平方差的因式分解公式[a^2-b^2=(a+b)(a-b)],先把四次方化为二次方,再按三角函数化简即可.cin^4α-sin^4α=(con^2α)^2-(sin^2α)^2=(con^2α+sin^2α)(con^2α-sin^2α)因sin^2α+con^2α=1,con^2α-sin^2...

cos^4a-sin^4a=(cos^2a-sin^2a)(cos^2a+sin^2a)
=cos^2a-sin^2a
=cos2a

tan（π/4+α）=(tanπ/4+tana)/(1-tanπ/4*tanα)

=-3