1.化简：（1-sin^6α -cos^6α)/(1-sin^4α -cos^4α)

题目

1.化简：（1-sin^6α -cos^6α)/(1-sin^4α -cos^4α)
2.α为三角形内角,lg(sinα + cosα)=1/2lg5 -lg3,求sinα - cosα,tanα的值.
3.已知：log(tanα)(cosα)=2/3 (0＜α＜π/2),求证:log(1+cot^2α)(sinα*cosα)=-7/10.
4.已知：5sinβ=sin（2α+β） ,求tan(α+β)*cotα 的值.

答案

零分?
①1-sin^6α -cos^6α
=1-（sin^2α+cos^2α）[(sin^2α+cos^2α）^2-3sin^2α*cos^2α）)
=3sin^2α*cos^2α
1-sin^4α -cos^4α=1-[(sin^2α+cos^2α）^2-2sin^2α*cos^2α]
=2sin^2α*cos^2α
原式=3/2
②lg[(sinα + cosα)^2]=lg5 -lg9=lg5/9
(sinα + cosα)^2=5/9=1+2sinαcosα
2sinαcosα=-4/9
1-2sinαcosα=(sinα - cosα)^2=13/9
sinα - cosα=±√13/3
sinα + cosα=√5/3
2sinα=√5/3±√13/3负数去掉
sinα=（√5+√13）/6
cosα=（√5-√13）/6
tanα=（√5+√13)/（√5-√13)=-(9+√65)/3
③看不懂log(tanα)(cosα)=2/3=log[(tanα)(cosα)]?
log(tanα)(cosα)=2/3=log(sinα)=2/3
sin^2α=(10)^4/3=10*10^(1/3)
cos^2α=1- (10)^4/3
log(1+cot^2α)(sinα*cosα)=-log(sin^3αcosα)
④α+β=θ,
5sin(θ-α)=sin(θ+α)
4sinθcosα=6cosθsinα
tan(α+β)*cotα =tanθcotα=2/3