Question

若sinθ+cosθ=k,且sin3θ+cos3θ<0,求k的取值范围

Answer 1

sin3θ+cos3θ
=sin2θcosθ+cos2θsinθ+cos2θcosθ-sin2θsinθ
=sin2θ(cosθ-sinθ)+cos2θ(cosθ+sinθ)
=2sinθcosθ(cosθ-sinθ)+(cosθ+sinθ)²(cosθ-sinθ)
=(cosθ-sinθ)[2sinθcosθ＋(cosθ＋sinθ)²]
=(cosθ+sinθ)[(cosθ+sinθ﹚²－1+(cosθ＋sinθ)²]
=(cosθ+sinθ)[2(cosθ+sinθ﹚²－1]
=k﹙2k²-1﹚＜0
∴k＜﹣√2/2或0＜k＜√2/2

Answer 2

解：已知：sinθ+cosθ=k (sinθ+cosθ)^2=k^2 sin2θ=k^2-1
(sinθ-cosθ)^2=1-sin2θ=2-k^2
sinθ-cosθ=(2-k^2)^1/2 或，sinθ+cosθ=-(2-k^2)^1/2，且-2^(1/2)<k<2^(1/2)
sin3θ+cos3θ
=sin2θcosθ+cos2θsinθ+cos2θcosθ-sin2θsinθ
=sin2θ(cosθ-sinθ)+cos2θ(cosθ+sinθ)
=2sinθcosθ(cosθ-sinθ)+(cosθ+sinθ)²(cosθ-sinθ)
=(cosθ-sinθ)[2sinθcosθ＋(cosθ＋sinθ)²]
=(cosθ-sinθ)(4sinθcosθ＋1)
=(cosθ-sinθ)(2sin2θ＋1)
（1）：当sinθ-cosθ=(2-k^2)^1/2
sin3θ+cos3θ=[(2-k^2)^1/2](2k^2-1)<0
解不等式：- √2/2 <k<√2/2
(2): 当sinθ-cosθ=-(2-k^2)^1/2
sin3θ+cos3θ=[-(2-k^2)^1/2](2k^2-1)<0
解不等式：-√2<k<-√2/2,或√2/2<k<√2.