已知函数f(x)=sin(x+π/12),x∈R(1)求f(-π/4)的值(2)若cosθ=4/5,θ∈(0,π/2),求f(2θ-π/3)

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已知函数f(x)=sin(x+π/12),x∈R(1)求f(-π/4)的值(2)若cosθ=4/5,θ∈(0,π/2),求f(2θ-π/3)

已知函数f(x)=sin(x+π/12),x∈R(1)求f(-π/4)的值(2)若cosθ=4/5,θ∈(0,π/2),求f(2θ-π/3)
已知函数f(x)=sin(x+π/12),x∈R
(1)求f(-π/4)的值
(2)若cosθ=4/5,θ∈(0,π/2),求f(2θ-π/3)

已知函数f(x)=sin(x+π/12),x∈R(1)求f(-π/4)的值(2)若cosθ=4/5,θ∈(0,π/2),求f(2θ-π/3)
(1)令X=-π/4,带入f(x)的方程有:
f(-π/4)=sin(-π/4+π/12)=sin(-π/6)=-sin(π/6)=-1/2
(2)∵cosθ=4/5,θ∈(0,π/2),∴sinθ=3/5,代入方程有:
f(2θ-π/3)=sin(2θ-π/3+π/12)=sin(2θ-π/4)=sin(2θ)cos(π/4)-cos(2θ)sin(π/4)=2sinθcosθcos(π/4)-(2cos²θ-1)sin(π/4)=(17√2)/50

问题一,将数代入,得到
sin(-π/6)=-1/2
问题二,代入得到sin(2θ-π/4)=sin2θ*cos-π/4—cos2θ*sinπ/4,sin2θ=2cosθ*sinθ=24/25,cos2θ=2cosθ*cosθ-1=7/25,
所以最终答案为17√2/50