已知函数f(x)=√2cos(x-π/12),x∈R

问题描述:

已知函数f(x)=√2cos(x-π/12),x∈R
求f(3/π)的值 2.若cosθ=3/5,θ€(3/2π,2π)求f(θ-π/6)

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