b(2cos^2ωx-1)應(yīng)該是b(2cos^ωx-1)吧
化簡得：asin2wx+bcos2wx=sqr(a^2+b^2)sin(2wx+φ)……(φ是設(shè)的)
所以a^2+b^2=4
2w=2π/(2π/2)=2
π/6+φ=π/2+2kπ,k∈Z
所以φ可以是π/3
所以a/sqr(a^2+b^2)=cosπ/3=0.5
所以 a=1
b=sqr(3)
(2). f(α)=2sin(2α+π/3)=2/3
所以sin(2α+π/3)=1/3
1-2sin^2(2α+π/3)=cos(4α+2π/3)=7/9
cos(4α+2π/3)=-cos(4α+2π/3-π)=-cos(4α-π/3)=-cos(π/3-4α)=-sin(π/3-4α+π/2)=-sin(5π/6-4α)=7/9
所以sin(5π/6-4α)=-7/9