三角函数及性质/21
若0<θ<π/2,求使(cos(θ))^2+2m×sin(θ)-2m-2<0恒成立的实数m的范围
解:(cos(θ))^2+2m×sin(θ)-2m-2 =-sin^θ+2msinθ-2m-1<0 △=4m^+4(2m-1)=m^+2m-1<0 ∴-1-√2<m<-1+√2 又0≤sinθ≤1 当0=sinθ时: m>-1/2 ∴-1/2<m<-1+√2
答:设sinx+cosx=(√2)sin(x+π/4) 设它等于 t (√2<t<√2) 则两边平方,有1+2sinxcosx=t^2 则 sinxcosx=(...详情>>