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)需要过程!
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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)需要过程!](/uploads/image/z/8503692-60-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28x%2B%CF%80%2F12%29%2Cx%E2%88%88R%2C%281%29%E6%B1%82f%28-%CF%80%2F4%29%E7%9A%84%E5%80%BC%2C%282%29%E8%8B%A5cos%CE%98%3D4%2F5%2C%CE%98%E2%88%88%280%2C%CF%80%2F2%29%2C%E6%B1%82f%282%CE%98-%CF%80%2F3%29%E9%9C%80%E8%A6%81%E8%BF%87%E7%A8%8B%21)
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)需要过程!
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)需要过程!
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)需要过程!
f(x)=sin(x+π/12)
(1) f(-π/4)=sin(-π/4+π/12)
=sin(-π/6)
=-1/2
(2) cosx=4/5
x∈(0,π/2)
sinx=√(1-(4/5)^2)
=3/5
f(2x-π/3)=sin(2x-π/3+π/12)
=sin(2x-π/4)
=sin2xcosπ/4-cos2xsinπ/4
=√2/2sinxcosx-√2/2(cos^2x-sin^2x)
=√2/2(3/5*4/5-(4/5)^2+(3/5)^2)
=√2/2(12/25-16/25+9/25)
=√2/10
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