tan(π/4+θ)+tan(π/4-θ)=4,cos2θ=?
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tan(π/4+θ)+tan(π/4-θ)=4,cos2θ=?
![tan(π/4+θ)+tan(π/4-θ)=4,cos2θ=?](/uploads/image/z/18990428-68-8.jpg?t=tan%28%CF%80%2F4%2B%CE%B8%29%2Btan%28%CF%80%2F4-%CE%B8%29%3D4%2Ccos2%CE%B8%3D%3F)
设π/4+θ=A,则θ=A-π/4,π/4-θ=π/2-A,cos2θ=cos(2A-π/2)=sin(2A).
tan(π/4+θ)+tan(π/4-θ)=tanA+tan(π/2-A)=tanA+cotA.
然后切化弦,再通分得原式=1/sinAcosA=2/sin2A=4
cos2θ=sin2A=1/2.
tan(π/4+θ)+tan(π/4-θ)=tanA+tan(π/2-A)=tanA+cotA.
然后切化弦,再通分得原式=1/sinAcosA=2/sin2A=4
cos2θ=sin2A=1/2.
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