Question

已知sinα-sinβ=-1/3,cosα-cosβ=1/2，α，β是锐角，求tan(α-β)

第二题 已知cos(α-β)=12/13, sin(α+β)=-3/5, π/2<β<α<3π/4,求sin2α
第三题 已知0<β<π/4<α<3π/4. cos(π/4-α)=3/5, sin(3π/4+β)=5/13,求sin(α+β)
求详细过程。。。谢谢

Answer 1

1、两式分别两边平方，相加，
（sinα)^2-2sinαsinβ+(sinβ)^2=1/9.
(cosα)^2-2cosαcosβ+(cosβ)^2=1/4,
1-2cos(α-β)+1=13/36,
cos(α-β)=59/72>0,α<β
α-β是第四象限的负角，
sin(α-β)<0,
sin(α-β)=-√{1-[cos(α-β)]^2}=-√1703/72，
tan(α-β)=（-√1703/72）/（59/72）=-√1703/59.
2、π<α+β<3π/2,
cos(α+β)<0,
cos(α+β)=-4/5,
0<α-β<π/2,
sin(α-β)>0,
sin(α-β)=5/13,
sin2α=sin(α-β+α+β)=sin(α+β)cos(α-β)+cos(α+β)sin(α-β)
=-56/65.
3、0<α-π/4 <π/2,
3π/4<3π/4+β<π,
cos(α-π/4)= cos(π/4-α)=3/5
sin(π/4-α)=4/5,
cos(3π/4+β)=-12/13,
sin(α+β）=sin(α-π/4+3π/4+β-π/2)
=-sin[π/2-(α-π/4+3π/4+β)]
=-cos(α-π/4+3π/4+β)
=-cos(α-π/4)cos(3π/4+β)+sin(α-π/4)sin(3π/4+β)
=-(3/5)*(-12/13)+(4/5)*(5/13)
=56/65.

追问

第二题 π<α+β<3π/2,和0<α-β<π/2怎么算。。。。。谢谢