已知f(x)=√2cos(x-π/12),x∈R(1)求f(-π/6)的值(2)若cosθ=3/5,θ∈(3π/2,2π),求f(2θ+π/3)

Question

已知f(x)=√2cos(x-π/12),x∈R (1)求f(-π/6)的值 (2)若cosθ=3/5,θ∈(3π/2,2π),求f(2θ+π/3)

Answer 1

（1）f(-π/6)=√2cos(-π/6-π/12)=√2cosπ/4=1
(2)∵cosθ=3/5,θ∈(3π/2,2π),
∴sinθ=-4/5
∴f(2θ+π/3)=√2cos(2θ+π/3-π/12)=√2cos(2θ+π/4)
=√2cos2θcosπ/4-√2sin2θsinπ/4
=cos2θ-sin2θ
=2cos²θ-1-2sinθcosθ
=2×9/25-1+2×4/5×3/5
=17/25