已知函数f(x)=根2cos(x-x/12),x∈R.求f(-π/6)的值.2)若cosy=3/5

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已知函数f(x)=根2cos(x-x/12),x∈R.求f(-π/6)的值.2)若cosy=3/5
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已知函数f(x)=根2cos(x-x/12),x∈R.求f(-π/6)的值.2)若cosy=3/5
已知函数f(x)=根2cos(x-x/12),x∈R.求f(-π/6)的值.2)若cosy=3/5

已知函数f(x)=根2cos(x-x/12),x∈R.求f(-π/6)的值.2)若cosy=3/5
已知函数f(x)=(√2)cos(x-π/12),x∈R;(原题的x/12可能是π/12之误)
(1).求f(-π/6)的值.
(2).若cosθ=3/5,θ∊(3π/2,2π),求f(2θ+π/3).
(1).f(-π/6)=(√2)cos(-π/6-π/12)=(√2)cos(-3π/12)=(√2)cos(π/4)=1
(2).θ∊(3π/2,2π),故2θ∊(3π,4π);cos2θ=2cos²θ-1=2(3/5)²-1=-7/25
故2θ是第二象限的角.于是sin2θ=√(1-49/625)=24/25.
∴f(2θ+π/3)=(√2)cos(2θ+π/3-π/12)=(√2)cos(2θ+π/4)
=cos2θ-sin2θ=-7/25-24/25=-31/25.