若3sinθ-cosθ=0,则(cos^2)θ+1/2sin2θ是?
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若3sinθ-cosθ=0,则(cos^2)θ+1/2sin2θ是?
![若3sinθ-cosθ=0,则(cos^2)θ+1/2sin2θ是?](/uploads/image/z/19418031-63-1.jpg?t=%E8%8B%A53sin%CE%B8-cos%CE%B8%3D0%2C%E5%88%99%EF%BC%88cos%5E2%EF%BC%89%CE%B8%2B1%2F2sin2%CE%B8%E6%98%AF%3F)
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
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
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