已知π⼀2小于β<α<3π⼀4,且cos (α-β)=12⼀13,sin(α+β)=-3⼀5,求cos2α的值,跪求解题过程和正确答案啊

2024-11-18 19:27:51
推荐回答(1个)
回答1:

π/2<β<α<3π/4,第二象限
π<α+β<3π/2第三象限
0<α-β<π/4第一象限

cos( α-β)=12/13==> sin( α-β)=5/13,
sin(α+β)=-3/5==>cos(α+β)=-4/5

sin2α
=sin(α+β + α-β)
=sin(α+β)cos(α-β)+cos(α+β)sin(α-β)
=-3/5*12/13 -4/5*5/13
=-56/65

cos2α
=cos(α+β + α-β)
= -sin(α+β)sin(α-β)+cos(α+β)cos(α-β)
=-(-3/5)*5/13 -4/5*12/13
=-33/65