已知函数f(x)=√2sin(x-π/12),x属于R,1.求f(-π/6)的值.2.若sinθ=-4/5,θ属于(3π/2,2π),求f(2θ+π/3)
问题描述：
已知函数f(x)=√2sin(x-π/12),x属于R,1.求f(-π/6)的值.2.若sinθ=-4/5,θ属于(3π/2,2π),求f(2θ+π/3)
答
1、f(-π/6)=√2sin(-π/6-π/12)
=√2sin(-π/4)
=√2*(-√2/2)
=-1
2、因为sinθ=-4/5,θ属于(3π/2,2π)
所以cosθ=3/5
f(2θ+π/3)=√2sin(2θ+π/3-π/12)
=√2sin(2θ+π/4)
=√2(√2/2*sin2θ+√2/2*cos2θ)
=2sinθcosθ+cos^2θ-sin^2θ
=2*(-4/5)*(3/5)+9/25-16/25
=-24/25+9/25-16/25
=-31/25