数学三角函数化简题目

（1）sin^2·α+cos^4·α+sin^2·αcos^2·α
=sin^2·α+cos^2·α(cos^2·α+sin^2·α)
= sin^2·α+cos^2·α
=1

（2）（1-cos^4·α-sin^4·α）/(1-cos^6·α-sin^4·α)
=[(1+sin^2·α)(1-sin^2·α)-cos^4·α]/[(1+sin^2·α)(1-sin^2·α)-cos^6·α]
=[cos^2·α+cos^2·αsin^2·α-cos^4·α]/[cos^2·α+cos^2·αsin^2·α-cos^6·α]
=[cos^2·α(1-cos^2·α)+cos^2·αsin^2·α]/[cos^2·α(1-cos^4·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[cos^2·α(1-cos^2·α)(1+cos^2·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[cos^2·αsin^2·α(1+cos^2·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[2cos^2·αsin^2·α+sin^2·αcos^4·α]
=2/[2+cos^2·α]

1.=sin^2a-(1-sin^2a)^2+sin^2acos^2a=sin^2a+(1-sina^2)^2+sin^2a(1-sin^2a)=sin^2a+(1-sin^2a)=1
2.=[1-(cos^2a+sin^2a)^2+2sin^2acos^2a]/{[1-(cos^2a)^3]-sin^4a}
=[2sin^2acos^2a]/[(1-cos^2a)(1+cos^4a+cos^2a)-(sin^2)^2]
=[2sin^2acos^2a]/[(sin^2a)(1+cos^4a+cos^2a)-(sin^2)^2]
=[2sin^2acos^2a]/[(sin^2a)(1+cos^4a+cos^2a-sin^2a)]
=[2sin^2acos^2a]/[(1-cos^2a)(1+cos^4a+cos^2a)-(sin