Question

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)

Answer 1

1、f(π/3)=√2cos(x-π/12）
=√2cos(π/3-π/12）
=√2cos(π/4)
=1
2、∵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)
=cosθ+sinθ
=3/5-4/5
=-1/5

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追问

=√2cos(π/3-π/12）
=√2cos(π/4)为什么等于π/4啊?

追答

π/3-π/12
=4π/12-π/12
=3π/12
=π/4

Answer 2

f(x)=√2cos(x－π/12)

则：

f(π/3)=√2cos(π/3－π/12)=√2cos(π/4)=1

cosθ=3/5、θ∈(3π/2，2π)，则：sinθ=－4/5

则：

f(θ－π/6)=√2cos(θ－π/4)=√2×[(√2/2)cosθ＋(√2/2)sinθ]=cosθ＋sinθ=－1/5

Answer 3

解：
f(x)=√2cos(x-π/12）
f(π/3)=√2cos(π/3-π/12)
=√2cos(π/4)
=√2*√2/2=1

cosθ=3/5，θ€(3/2π,2π)
∴sinθ=-4/5
f(θ-π/6)=√2cos(θ-π/6-π/12)
=√2cos(θ-π/4)
=√2cosθcosπ/4+√2sinθsinπ/4
=√2*3/5*√2/2+√2*(-4/5)*√2/2
=3/5-4/5
=-1/5

Answer 4

f(π/3）=根号2cos（π/3-π/12)=根号2cosπ/4=1