已知cos(x-π/4)=√2/10,x∈(π/2,3π/4),1.求sinx 2.求sin(2x+π/3)的值
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1)
(∵x∈(π/2,3π/4),∴x-π/4∈(π/4,π/2)
∵cos(x-π/4)=√2/10
∴sin(x-π/4)=√[1-cos²(x-π/)]=7√2/10
∴sinx=sin[(x-π/4)+π/4]
=sin(x-π/4)cosπ/4+cos(x-π/4)sinπ/4
=7√2/10*√2/2+√2/10*√2/2
=4/5
(2)
cosx=-3/5
∴sin2x=2sinxcosx=2*4/5*(-3/5)=-24/25
cos2x=1-2sin²x=-7/25
∴sin(2x+π/3)=sin2xcosπ/3+cos2xsinπ/3
=-24/25*1/2-7/25*√3/2
=-(24+7√3)/50
(∵x∈(π/2,3π/4),∴x-π/4∈(π/4,π/2)
∵cos(x-π/4)=√2/10
∴sin(x-π/4)=√[1-cos²(x-π/)]=7√2/10
∴sinx=sin[(x-π/4)+π/4]
=sin(x-π/4)cosπ/4+cos(x-π/4)sinπ/4
=7√2/10*√2/2+√2/10*√2/2
=4/5
(2)
cosx=-3/5
∴sin2x=2sinxcosx=2*4/5*(-3/5)=-24/25
cos2x=1-2sin²x=-7/25
∴sin(2x+π/3)=sin2xcosπ/3+cos2xsinπ/3
=-24/25*1/2-7/25*√3/2
=-(24+7√3)/50
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