已知函数f(x)=2cos(x/4+π/6）,x∈R,且f(π/3)=√2设α,β∈［0,π/2］,f(4α+4π/3)=-30/17,f(4β-2π/3）=8/5求cosα+β值

Question

已知函数f(x)=2cos(x/4+π/6）,x∈R,且f(π/3)=√2
设α,β∈［0,π/2］,f(4α+4π/3)=-30/17,f(4β-2π/3）=8/5 求cosα+β值

Answer 1

f(x)=2cos(x/4+π/6）
f(4α+4π/3)=-30/17,那么
f(4α+4π/3）=2cos[(4α+4π/3)/4+π/6]
=2cos(α+π/2)=-2sinα=-30/17
∴sinα=15/17
∵α∈［0,π/2］,
∴cosα=√(1-sin²α)=8/17
f(4β-2π/3）=8/5
那么
2cos[(4β-2π/3)/4+π/6]=2cosβ=8/5
∴cosβ=4/5
∵β∈［0,π/2］
∴sinβ=√(1-cos²β)=3/5
∴cos(α+β)
=cosαcosβ-sinαsinβ
=8/17*4/5-15/17*3/5
=-13/85