Question

2、已知函数f(x)=√3sin2ωx-2sin^2ωx的最小正周期为3π.

⑴求函数f(x)的解析式
⑵在△ABC中,若f(C)=1,且2sin^2B=cosB+cos(A-C),求sinA的值

Answer 1

⑴f(x)=√3sin2ωx+cos2ωx-1=2sin(2ωx+π/6)-1 2π/2ω=3π→f(x)=2sin(2/3x+π/6)-1
⑵△ABC中,若f(C)=2sin(2/3C+π/6)-1=1,且2sin^2B=cosB+cos(A-C)→sin(2/3C+π/6),=1 且
1-cos2B =cosB+cos(A-C)→2/3C+π/6=π/2+2kπ C∈(0,π)→C=π/2
且1-cos2B =cosB+cos(A-π/2)=cosB+cosB=2cosB →1-2cos^2B+1=2-2cos^2B=2cosB→

cosB=(√5-1)/2→ sinA= sin(π/2-B)=cosB=(√5-1)/2