搜搜做题作业网已知函数f(x)=sin(x+π/12),x∈R,(1)求f(-π/4)的值,(2)若cosΘ=4/5,Θ∈(0,π/2
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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!
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 再答： 望采纳
(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 再答： 望采纳
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