已知函数f(x)=√2cos(x-π/12）,x∈R

1、f(π/3)=√2cos(x-π/12）=√2cos(π/3-π/12）=√2cos(π/4)=12、∵cosθ=3/5,θ€(3/2π,2π)∴sinθ=-4/5则f(θ-π/6)=√2cos(θ-π/6-π/12)=√2cos(θ-π/4)=√2cosθcos(π/4)+√2sinθsin(π/4)=cos...=√2cos(π/3-π/12） =√2cos(π/4)为什么等于π/4啊?π/3-π/12=4π/12-π/12=3π/12=π/4