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