求值已知x+y=3-cos2θ，
x+y=3-cos2θ，x-y=4sinθ，解得x=sinθ+1=[cos(θ/2)+sin(θ/2)]^ , y=sinθ-1=[cos(θ/2)-sin(θ/2)]^ , ∴ x＾(１/２)+y＾(１/２)=|cos(θ/2)+sin(θ/2)|+| cos(θ/2)- sin(θ/2)]
(1) 当θ∈[2kπ,2kπ+π/2]时, x＾(１/２)+y＾(１/２)=cos(θ/2)+sin(θ/2)+cos(θ/2)-sin(θ/2)=2cos(θ/2);
(2) 当θ∈[2kπ+π/2,2kπ+π]时, x＾(１/２)+y＾(１/２)=cos(θ/2)+sin(θ/2)+sin(θ...全部
x+y=3-cos2θ，x-y=4sinθ，解得x=sinθ+1=[cos(θ/2)+sin(θ/2)]^ , y=sinθ-1=[cos(θ/2)-sin(θ/2)]^ , ∴ x＾(１/２)+y＾(１/２)=|cos(θ/2)+sin(θ/2)|+| cos(θ/2)- sin(θ/2)]
(1) 当θ∈[2kπ,2kπ+π/2]时, x＾(１/２)+y＾(１/２)=cos(θ/2)+sin(θ/2)+cos(θ/2)-sin(θ/2)=2cos(θ/2);
(2) 当θ∈[2kπ+π/2,2kπ+π]时, x＾(１/２)+y＾(１/２)=cos(θ/2)+sin(θ/2)+sin(θ/2)-cos(θ/2)=2sin(θ/2);
(3) 当θ∈[2kπ+π,2kπ+3π/2]时, x＾(１/２)+y＾(１/２)=-cos(θ/2)-sin(θ/2)-cos(θ/2)+sin(θ/2)=-2cos(θ/2);
(4) 当θ∈[2kπ+3π/2,2kπ+2π]时, x＾(１/２)+y＾(１/２)=-cos(θ/2)+sin(θ/2)+cos(θ/2)+sin(θ/2)=-2sin(θ/2)。
(以上K∈Z)
。收起
