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)
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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)
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)
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)
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
f(x)=√2cos(x-π/12)
则:
f(π/3)=√2cos(π/3-π/12)=√2cos(π/4)=1
cosθ=3/5、θ∈(3π/2,2π),则:sinθ=-4/5
则:
f(θ-π/6)=√2cos(θ-π/4)=√2×[(√2/2)cosθ+(√2/2)sinθ]=cosθ+sinθ=-1/5
f(x)=√2cos(x-π/12)
f(π/3)=√2cos(π/3-π/12)
=√2cos(π/4)
=√2*√2/2=1
cosθ=3/5,θ€(3/2π,2π)
∴sinθ=-4/5
f(θ-π/6)=√2cos(θ-π/6-π/12)
=√2c...
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f(x)=√2cos(x-π/12)
f(π/3)=√2cos(π/3-π/12)
=√2cos(π/4)
=√2*√2/2=1
cosθ=3/5,θ€(3/2π,2π)
∴sinθ=-4/5
f(θ-π/6)=√2cos(θ-π/6-π/12)
=√2cos(θ-π/4)
=√2cosθcosπ/4+√2sinθsinπ/4
=√2*3/5*√2/2+√2*(-4/5)*√2/2
=3/5-4/5
=-1/5
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f(π/3)=根号2cos(π/3-π/12)=根号2cosπ/4=1