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已知tAn2α=3/4,α∈(%π/2,π/2),F(x)=sin(x+α)+...

那个,sin角不应该是负数的呀╭(°A°`)╮

∵ tan 2 α= 3 4 ∴tanα=± 3 2 ,∴sinα=± 21 7 f(x)=sin(x+α)+sin(α-x)-2sinα=2sinαcosx-2sinα=2sinα(cosx-1)当函数f(x)的最小值为0时,sinα<0,∴sinα=- 21 7 ∴cos2α=1-2sin 2 α=1-2× (- 21 7 ) 2 = 1 7 ∵sinα=- 21 7 ∴cosα= 2 7 7 ∴t...

∵tan2α=34∴tanα=±32,∴sinα=±217f(x)=sin(x+α)+sin(α-x)-2sinα=2sinαcosx-2sinα=2sinα(cosx-1)当函数f(x)的最小值为0时,sinα<0,∴sinα=-217∴cos2α=1-2sin2α=1-2×(?217)2=17∵sinα=-217∴cosα=277∴tanα2=sinα1+cosα=

等于60度,望采纳

(1)∵ tan2α= 2tanα 1- tan 2 α = 2( 2 -1) 1- ( 2 -1) 2 =1 又∵α为锐角∴α= π 8 ∴ sin(2α+ π 4 )=1 ∴f(x)=x 2 +x.(2)∵数列{a n }的首项 a 1 = 1 2 ,a n+1 =f(a n ),∴ a 2 = 1 4 + 1 2 = 3 4 , a 3 = 9 16 + 3 4 = 21 16 , a 4 = 44...

答: 因为:π/2

∵tan2α= 2tanα 1-ta n 2 α = 2 2 -2 1- ( 2 -1) 2 =1又∵α为锐角∴α= π 8 ∴sin(2α+ π 4 )=1∴f(x)=x 2 +xf(-1)=(-1) 2 -1=0.故答案为:0.

f(x)=sin(x+θ)+sin(x-θ)-2sinθ =2[sin(x+θ+x-θ)/2*cos(x+θ-x+θ)/2]-2sinθ =2sinx*cosθ-2sinθ=√(4sin^2x+4)cos(α+θ)≥0 θ∈(0,π/2) tan2θ=-3/4 2θ∈(π/2,π) sin2θ=-3/4cos2θ (sin2θ)^2+(cos2θ)^2=25(cos2θ)^2/16=1 cos2θ=-4/5 cos2θ=2(cosθ)^2...

f(x)=sin(θ+x)+sin(θ-x)-2sinθ=2sinθcosx-2sinθ =2sinθ(cosx-1) x∈R cosx-1≤0 要满足题意,则sinθ≤0 θ在第三、四象限 又θ∈(0,3π/2),所以θ∈(π,3π/2),cosθ≤0 2θ∈(2π,3π),tan2θ=-3/4<0 2θ∈(5π/2,3π), cos2θ<0 cos2θ= -4/5 cosθ= -√[(...

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