设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .(2) 若f(β)=2,β∈[0,2π],求β
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![设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .(2) 若f(β)=2,β∈[0,2π],求β](/uploads/image/z/3779439-15-9.jpg?t=%E8%AE%BEf%28x%29%3Dtanx%281%2Bsin2x%2Bcos2x%29%2C%EF%BC%881%EF%BC%89+tan%28%CE%B1%2B%CF%80%2F4%29%3D2%2C%E6%B1%82f%28%CE%B1%29+.%E8%AE%BEf%28x%29%3Dtanx%281%2Bsin2x%2Bcos2x%29%2C%EF%BC%881%EF%BC%89+tan%28%CE%B1%2B%CF%80%2F4%29%3D2%2C%E6%B1%82f%28%CE%B1%29+.%EF%BC%882%EF%BC%89+%E8%8B%A5f%28%CE%B2%29%3D2%2C%CE%B2%E2%88%88%5B0%2C2%CF%80%5D%2C%E6%B1%82%CE%B2)
设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .(2) 若f(β)=2,β∈[0,2π],求β
设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .
设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .
(2) 若f(β)=2,β∈[0,2π],求β
设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .设f(x)=tanx(1+sin2x+cos2x),(1) tan(α+π/4)=2,求f(α) .(2) 若f(β)=2,β∈[0,2π],求β
f(x)=tanx(1+sin2x+cos2x)
=tanx(1+2sinxcosx+2cos²x-1)
=tanx(2sinxcosx+2cos²x)
=2sin²x+2sinxcosx
(1) tan(α+π/4)=2
tan(α+π/4)
=(tanα+1)/(1-tanα)=2
∴tanα=1/3
f(α)=2sin²α+2sinαcosα
=(2sin²α+2sinαcosα)/1
=(2sin²α+2sinαcosα)/(sin²α+cos²α)
分子分母同除cos²α得
=(2tan²α+2tanα)/(tan²α+1)
=4/5
(2)
f(β)=2,β∈[0,2π],
2sin²β+2sinβcosβ=2
sin²β+sinβcosβ=1
∵sin²β+cos²β=1
∴sinβcosβ=cos²β
∴cosβ(sinβ-cosβ)=0
β=π/2,3π/2
或β=π/4,5π/4