Question

已知函数f(x)=√2cos(x-π/12），xR∈

求f(-π/6)的值 2.若cosθ=3/5，θ€(3/2π,2π)求f(2θ+π/3)

Answer 1

f(x)=√2cos(x-π/12）
f(-π/6) = √2cos(-π/6-π/12）= √2cos(-π/4) = √2cos(π/4) = √2*√2/2 = 1

f(2θ+π/3)= √2cos(2θ+π/3-π/12) = √2cos(2θ+π/4)

θ属于(3/2π，2π)

2θ属于(3π，4π)
2θ+π/4属于(3π+π/4，4π+π/4)
2θ=3π+π/4时有最小值√2cos(3π+π/4) = √2*(-√2/2) = -1
2θ=4π时有最大值√2cos(4π) = √2*1 = √2

Answer 2

1)f(-π/6)= √2cos(-π/6-π/12）=√2cos(-π/4)=1

2)cosθ=3/5，θ€(3/2π,2π)，则sinθ=-√(1-cosθ*cosθ)=-4/5
f(2θ+π/3)=√2cos(2θ+π/3-π/12）=√2cos(2θ+π/4）=cos2θ-sin2θ=2cosθ*cosθ-1-2cosθ*sinθ=17/25