RT~三题1.在△ABC中,(1)若sin（B+C/2）=4/5,求cos（A-B）的值；(2),若sinAsinB=c

A+C/2+B+C/2=π,A+C/2=π-B-C/2
sin（A+C/2）=sin（π-B-C/2）=sin（B+C/2）=4/5
cos（A+C/2）=cos（π-B-C/2）=-cos（B+C/2）
[cos(B+C/2)]^2=1-(4/5)^2=(3/5)^2
cos（A-B）=cos[(A+C/2)-(B+C/2)]
=cos（A+C/2）cos（B+C/2)+sin（A+C/2）sin（B+C/2）
=-cos(B+C/2)cos(B+C/2)+sin（A+C/2）sin（B+C/2）
=-(3/5)^2+(4/5)(4/5)
=7/25
sinAsinB=cos(平方)C/2
sinAsinB=(cosC+1)/2
2sinAsinB=cosC+1
2sinAsinB=-cos(A+B)+1
2sinAsinB=-cosAcosB+sinAsinB+1
cosAcosB+sinAsinB=1
cos(A-B)=1
因为A,B为三角形的内角,所以A-B=0
A=B,△ABC为等腰三角形
2.8cos(π/4+α)cos(π/4-α)=1
8[cos(π/4)cosα-sin(π/4)sinα][cos(π/4)cosα+sin(π/4)sinα]=1
8[(根号2)/2](cosα-sinα)[(根号2)/2](cosα+sinα)=1
4[(cosα)^2-(sinα)^2]=1
(cosα)^2-(sinα)^2=1/4
(cosα)^2+(sinα)^2=1
(cosα)^2=5/8,(sinα)^2=3/8
sin(四次方)α+cos(四次方)α
=[(cosα)^2+(sinα)^2]^2-2[(cosα)^2][(sinα)^2]
=1-2*5/8*3/8
=17/32
F(θ)=cos(平方)(θ+α)+cos(平方)(θ+β)
=[cos(2θ+2α)+1]/2+[cos(2θ+2β)+1]/2
=[cos(2θ+2α)+cos(2θ+2β)]/2+1
要使F(θ)的值不随θ的变化而变化
即cos(2θ+2α)+cos(2θ+2β)不随θ的变化而变化
cos(2θ+2α)+cos(2θ+2β)
=cos2θcos2α-sin2θsin2α+cos2θcos2β-sin2θsin2β
=cos2θ(cos2α+cos2β)-sin2θ(sin2α+sin2β)不随θ的变化而变化
所以cos2α+cos2β=0,且sin2α+sin2β=0
cos2α+cos2β=0
2(cosα)^2-1+1-2(sinβ)^2=0
(cosα)^2-(sinβ)^2=0
(cosα)^2=(sinβ)^2
cosα=0时,
0≤α＜β≤π,α=π/2
sinβ=0,β≤π
β-α=π/2
cosα 不等于 0时,sinβ亦不等于 0
sin2α+sin2β=0
2sinαcosα+2sinβcosβ=0
sinαcosα+sinβcosβ=0,两边乘cosα
得 cosαsinαcosα+cosαsinβcosβ=0
sinα(sinβ)^2+cosαsinβcosβ=0,两边除sinβ
得 sinβsinα+cosαcosβ=0
cos(α-β)=0
0≤α＜β≤π,所以-π≤α-β