1.
sinα+sinβ=√2(平方)
(sinα+sinβ)^2=2
(sinα)^2+2sinαsinβ+(sinβ)^2=2.1
cosα+cosβ=2√3/3(平方)
(cosα+cosβ)^2=4/3
(cosα)^2+2cosαcosβ+(cosβ)^2=4/3.2
1式+2式得
2cosαcosβ+2sinαsinβ=10/3
cosαcosβ+sinαsinβ=5/3
cos(α-β)=5/3
2.题目有误
[(sinα)^2+tanα*tan(α/2)+cosα) ^2]*sin2α/2cosα
=[1+tanα*tan(α/2)]*sin2α/2cosα
={1+2tan(α/2)/[1-tan^2(α/2)]*tan(α/2)}*sin2α/2cosα
={1+2tan^2(α/2)/[1-tan^2(α/2)]}*sin2α/2cosα
=[1+tan^2(α/2)]/[1-tan^2(α/2)]*sin2α/2cosα
=[sec(α/2)]^2 /[1-tan^2(α/2)]*sin2α/2cosα
=[sec(α/2)]^2 /{2-[sec(α/2)]^2)]*sin2α/2cosα
=1/[2/sec^2(α/2)-1]*sin2α/2cosα
=1/[2cos^2(α/2)-1]*sin2α/2cosα
=1/cosα*sin2α/2cosα
=1/cosα*2inαcosα/2cosα
=snα/cosα
=tanα.
3.
tanα/2=2
tanα=2tanα/2/[1-tan^2α/2]
=2*2/[1-4]
=-4/3
(6sinα+cosα)/(3sinα-2cosα)上下同除cosα
=(6tanα+1)/(3tanα-2)
=[6*(-4/3)+1]/[3*(-4/3)-2]
=7/6
4.
(cosα)^2/[cot(α/2)-tan(α/2)]
=(cosα)^2 /[(1+cosα)/sinα]-[(1-cosα)/sinα]
=(cosα)^2/(2cosα/sinα)
=1/2 *sinαcosα
=1/4 *2sinαcosα
=1/4sin2α
注：(1/cotθ/2)=tanθ/2=sinθ/(1+cosθ)=(1-cosθ)/sinθ
5.
(sin2α)＾2+sin2αcosα-cos2α=1
4(sinα)＾2*(cosα)＾2+2sinα*(cosα)＾2-[2(cosα)＾2-1]=1
4(sinα)＾2*(cosα)＾2+2sinα*(cosα)＾2-2(cosα)＾2=0
因为 α∈(0,π/2),所以cosα≠0
所以2(sinα)＾2+sinα-1=0
(2sinα-1)*(sinα+1)=0
(sinα)1=1/2,(sinα)2=-1[舍去,α∈(0,π/2),sinα>0]
所以sinα=1/2,
α=π/6
cosα=√3/2
tanα=sinα/cosα=√3/3
