已知fx=根号2cos(x-π/2),x∈R求f（-π/3）的值若sinθ=-3/5,θ∈[3π/2,2π],求f（2θ+π/3）的值

Question

已知fx=根号2cos(x-π/2),x∈R

求f（-π/3）的值

若sinθ=-3/5,θ∈[3π/2,2π],求f（2θ+π/3）的值

Answer 1

f(x)=√2cos(x-π/2)=√2sinx
(1)f(-π/3)=√2sin(-π/3)=-√6/2
(2)∵θ∈[3π/2,2π]
∴cosθ=4/5
∴sin2θ=2sinθcosθ=-24/25 cos2θ=1-2sin^2θ=7/25
f(2θ+π/3)=√2sin(2θ+π/3)=√2(sin2θ/2+√3cos2θ/2)=2√6/50-12√2/25