三角函数求解

Question

如题所述

Answer 1

(1)5π/12-π/6=π/4=T/4, T=π, 所以,ω=2, f(x)=2sin(2x+ϕ),把点(π/6,2)代入，2sin(2*π/6+ϕ)=2, sin(π/3+ϕ)=1, π/3+ϕ=π/2, ϕ=π/6
所以y=2sin(2x+π/6)

(2)f(B)=2sin(2B+π/6)=1, sin(2B+π/6)=1/2, 2B+π/6=π/6或5π/6, 所以B=π/3
(a+c)/b=(sinA+sinC)/sinB=[2sin(A+C)/2cos(A-C)/2]/sinB
分子=2sin[(A+C)/2]cos[(A-C)/2], A+C=π-B=2π/3, (A+C)/2=π/3, 所以分子中的sin[(A+C)/2]=sin(π/3), 可与分母约去, 所以(a+c)/b=2cos[(A-C)/2]
A+C=2π/3, C=(2π/3)-A, A-C=A-[(2π/3)-A]=2A-2π/3 (A-C)/2=A-(π/3)
0<A<2π/3, -π/3<A-(π/3)<π/3,
1/2<cos[(A-C)/2]≤1, 因为(a+c)/b=2cos[(A-C)/2],所以 1<(a+c)/b≤2追问

谢谢