Question

已知π/2＜β＜a＜3π/4，cos(a-π)=12/13，sin(a+β)=-3/5，求sin2β和cos2a

Answer 1

∵π/2＜β＜a＜3π/4，cos(a-π)=12/13
∴cosa=-12/13,sina=5/13
cos2a=2cosa*cosa-1=119/169
又sin(a+β)=-3/5<0,且π/2＜β＜a＜3π/4
∴a+β在第三象限，即cos(a+β)=-4/5
sinβ=sin(a+β-a)=sin(a+β)cosa-cos(a+β)sina=56/65
则cosβ=-33/65
sin2β=2sinβcosβ=-3696/4225

Answer 2

因为cos（a-b）=12/13,所以sin(a-b)=5/13
sin(a+b)=-3/5(π<a+b<3/2π) 所以cos(a+b)=-4/5,
cos2a=cos{(a+b)+（a-b)}=cos(a+b)cos(a-b)-sin(a-b)sin(a+b)=-4/5*12/13+5/13*3/5=-33/65
sin2b=sin{(a+b)-(a-b)}=sin(a+b)cos(a-b)-sin(a-b)cos(a+b)=-3/5*12/13+5/13*4/5=-16/65