Question

已知函数fx=sin(x+π/12)求f(-π/4)

1.已知函数fx=sin(x+π/12)求f(-π/4)
2.若cosθ=4/5 θ∈（0，π/2）求2θ-π/3
.若cosθ=4/5 θ∈（0，π/2）求f（2θ-π/3）

Answer 1

1、f（-π/4)=sin(-π/4+π/12)=sin(-π/6)=-1/2

追问

恩对的，下面还有一个问呢若cosθ=4/5 θ∈（0，π/2）求f（2θ-π/3）

追答

cosθ=4/5。sinθ=3/5
sin2θ=2sinθcosθ=12/25
cos2θ=√13/25
f(x)=sin(2θ-π/3+π/12)=sin(2θ-π/4)
=sin2θcosπ/4-cos2θsinπ/4
=(12/25)(√2/2)-(√13/25)(√2/2)
=(12√2-√26）/25

Answer 2

1. f(-π/4)=sin(-π/4+π/12)=sin(-π/6)=-1/2

2. cosθ=4/5, θ=36.67度， 2θ-π/3=13.74度

3. cosθ=4/5,sinθ=3/5,sin2θ=24/25, cos2θ=7/25

   f（2θ-π/3）=cos(2θ-π/3）=1/2cos2θ+√3/2sin2θ=(7+24√3)/50