已知函数f(x)=√2cos(x-π/12),x∈R
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已知函数f(x)=√2cos(x-π/12),x∈R
求f(3/π)的值 2.若cosθ=3/5,θ€(3/2π,2π)求f(θ-π/6)
求f(3/π)的值 2.若cosθ=3/5,θ€(3/2π,2π)求f(θ-π/6)
1、f(π/3)=√2cos(x-π/12)
=√2cos(π/3-π/12)
=√2cos(π/4)
=1
2、∵cosθ=3/5,θ€(3/2π,2π)
∴sinθ=-4/5
则f(θ-π/6)
=√2cos(θ-π/6-π/12)
=√2cos(θ-π/4)
=√2cosθcos(π/4)+√2sinθsin(π/4)
=cosθ+sinθ
=3/5-4/5
=-1/5
再问: =√2cos(π/3-π/12) =√2cos(π/4)为什么等于π/4啊?
再答: π/3-π/12 =4π/12-π/12 =3π/12 =π/4
=√2cos(π/3-π/12)
=√2cos(π/4)
=1
2、∵cosθ=3/5,θ€(3/2π,2π)
∴sinθ=-4/5
则f(θ-π/6)
=√2cos(θ-π/6-π/12)
=√2cos(θ-π/4)
=√2cosθcos(π/4)+√2sinθsin(π/4)
=cosθ+sinθ
=3/5-4/5
=-1/5
再问: =√2cos(π/3-π/12) =√2cos(π/4)为什么等于π/4啊?
再答: π/3-π/12 =4π/12-π/12 =3π/12 =π/4
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