设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π Bπ/4≤x≤7π/4 Cπ/4≤x≤5π/4 Dπ/2≤x(1)设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π B.π/4≤x≤7π/4 C.π/4≤x≤5π/4 D.π/2≤x≤3π/2(2)已
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设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π Bπ/4≤x≤7π/4 Cπ/4≤x≤5π/4 Dπ/2≤x(1)设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π B.π/4≤x≤7π/4 C.π/4≤x≤5π/4 D.π/2≤x≤3π/2(2)已
设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π Bπ/4≤x≤7π/4 Cπ/4≤x≤5π/4 Dπ/2≤x
(1)设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则
A.0≤x≤π
B.π/4≤x≤7π/4
C.π/4≤x≤5π/4
D.π/2≤x≤3π/2
(2)已知sinθ+cosθ=1/5,θ∈(0,π)求sinθ,cosθ的值.
设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π Bπ/4≤x≤7π/4 Cπ/4≤x≤5π/4 Dπ/2≤x(1)设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π B.π/4≤x≤7π/4 C.π/4≤x≤5π/4 D.π/2≤x≤3π/2(2)已
C
θ在第一象限
得sinθ>0,cosθ>0
θ在第二象限
得sinθ>0,cosθ
1,C,1-2sinxcosx=(sinx-cosx)(sinx-cosx)=sinx-cosx,因而有sinx=cosx或sinx-cosx=1;即x=π/4,5π/4;或x=π/2
2,sin=4/5,cos=-3/5
(1) 展开后应是sin(x/2)=cos(x/2)则x就该为π/4
(2) 就是把sinθ+cosθ=1/5平方后得到2sinθcosθ=-24/25 再用换元法就可以解出来了
(1)C