Question

一道数学三角函数的题目

求详细过程
第28题

Answer 1

f(x)=sinωx-√3cosωx+1=2[sinωx*(1/2)-cosωx*(√3/2)]+1
=2sin[ωx-(π/3)]+1
周期T=2π/ω=6π
所以，ω=1/3
则，f(x)=2sin[(x/3)-(π/3)]+1
那么，f[3α-(π/2)]=2sin[α-(π/2)]+1=-2cosα+1=1/17
所以，cosα=8/17
那么，sinα=15/17
f(3β+π)=2sinβ+1=11/5
所以，sinβ=3/5
所以，cosβ=4/5
则，cos(α+β)=cosαcosβ-sinαsinβ=(8/17)*(4/5)-(15/17)*(3/5)=-13/85

Answer 2

(1) f(x)=sinωx-√3cosωx+1
=2sin(ωx-π/3)+1
T=6π
2π/ω=6π
ω=1/3
(2) f(x)=2sin(x/3-π/3)+1
f(3α-π/2)=1/17
2sin((3α-π/2)/3-π/3)+1=1/17
2sin(α-π/2)+1=1/17
-2sinα=-16/17
sinα=8/17
∵α∈[0,π/2]
∴cosα=√(1-sin^2α)
=15/17
f(3β+π)=11/5
2sin((3β+π)/3-π/3)+1=11/5
2sinβ+1=11/5
sinβ=3/5
∵β∈[0,π/2]
∴cosβ=√(1-sin^2β)
=4/5
cos(α+β)=cosαcosβ-sinαsinβ
=15/17*4/5-8/17*3/5
=36/85