已知f(x)=1/2sin2x+根号3/2cos2x-根号3/2求f(-25π/6)的值

问题描述:

已知f(x)=1/2sin2x+根号3/2cos2x-根号3/2求f(-25π/6)的值

变形:f(x)=sin(2x+π/3)-根号3/2 f(x)的周期为kπ(f(x+π)=f(x))
f(-25π/6)=f(-4π-π/6)=f(-π/6)=sin(-π/3+π/3)-根号3/2= -根号3/2.本题的关键是计算周期,从而简化计算!