若3sinθ-cosθ=0,则（cos^2）θ+1/2sin2θ是?
问题描述：
若3sinθ-cosθ=0,则（cos^2）θ+1/2sin2θ是?
答
3sinθ-cosθ=0
tanθ=1/3
(cosθ)^2+(1/2)sin(2θ)=(1/2)[cos(2θ)+1]+(1/2)sin(2θ)
=(1/2)+(1/2){[1-(tanθ)^2]/[1+(tanθ)^2]+2tanθ/[ 1+(tanθ)^2]}
=(1/2)+(1/2){[1-(tanθ)^2+2tanθ]/[ 1+(tanθ)^2]}
=(1/2){[1+(tanθ)^2+1-(tanθ)^2+2tanθ]/[ 1+(tanθ)^2]}
=[1+tanθ]/[ 1+(tanθ)^2]
=[1+1/3]/[ 1+1/9]
=6/5