分类：教育/科学-学习帮助
问题：三角证明题设xtan(θ
设xtan(θ-β)＝ytan(θ+β),(x-y)c2θ+(x+y)cos2β＝z.求证：2z(x+y)cos2β＝z^2+4xy.
回答：由xtan(θ-β)＝ytan(θ+β)
得:xsin(θ-β)cosx(θ+β)＝ysin(θ+β)cos(θ-β)
即xsin2θ-xsin2β=ysin2θ+ysin2β
即(x-y)sin2θ=(x+y)sin2β
又[z-(x+y)cos2β]^2=[(x-y)cos2θ]^2
即z^2-2z(x+y)cos2β=(x-y)^2-[(x-y)sin2θ]^2+[(x+y)sin2β]^2-(x+y)^2=(x-y)^2-(x+y)^2=-4xy
即2z(x+y)cos2β＝z^2+4xy
证毕.
(用手机回答真的好累啊！)