神马品牌网日本出名av女演员:三角函数问题.
来源：百度文库 编辑：神马品牌网 时间：2024/05/04 19:46:10
已知sin(x+y)cosz=-1,求证:(1)sin(2x+y)-siny=0;(2)tan(1/2(x+y+z))=-1.
(1)|sin(x+y)|≤1,|cosz|≤1
∴|sin(x+y)|×|cosz|≤1
sin(x+y)cosz=-1,∴sin(x+y)=1 或-1
1.sin(x+y)=1,x+y=π/2+2kπ ∴cosx=siny
sin(2x+y)=sin[x+(x+y)]=sinxcos(x+y)+sin(x+y)cosx=cosx=siny
∴sin(2x+y)-siny=0
类似的当sin(x+y)=-1 也成立
(2)sin(x+y)=1时,有cosz=-1,z=π+2kπ
∴0.5(x+y+z)=(3/4)π+2kπ
tan(1/2(x+y+z))=-1
同理,sin(x+y)=-1 cosz=1
tan(1/2(x+y+z))=-1