f(x)=sinxcosθ+cosxsinθ+sinθcosx-cosθsinx-2sinθ
=2sinθcosx-2sinθ
=2sinθ(1-cosx)
>=0
因为1-cosx>=0
所以sinθ>=0,又
tanθ=3/4>0
θ（0,3π/2]所以θ在（0,π/2）θ/2在（0,π/4)
cosθ=4/5
又cos²θ/2=1/2(1+cosθ)=1/2(1+4/5)=9/10
所以
cosθ/2=3v10/10cosθ符号没错吗？没错，sinθ>=0,又tanθ=3/4>0 θ（0,3π/2]
所以θ在（0，π/2）θ/2在（0，π/4)