问题求cos2α
已知cos(α+β)=4/5, cos(α-β).=-4/5,3π /2＜α+β＜2π, π/2＜α-β＜7,求cos2α.
已知cos(α+β)=4/5, cos(α-β).=-4/5,3π /2＜α+β＜2π, π/2＜α-β＜7,求cos2α.
答案sin(a+b)=-3/5,sin(a-b)=3/5
cos[(a+b)+(a-b)]=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
=4/5*(-4/5)-(-3/5)*3/5
=-7/25
即cos2a=-7/25
cos[(a+b)+(a-b)]=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
=4/5*(-4/5)-(-3/5)*3/5
=-7/25
即cos2a=-7/25
