sin(x+π/4)=5/13,x∈(π/2,3π/4),求sinx

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sin(x+π/4)=5/13,x∈(π/2,3π/4),求sinx

sin(x+π/4)=5/13,x∈(π/2,3π/4),求sinx
sin(x+π/4)=5/13,x∈(π/2,3π/4),求sinx

sin(x+π/4)=5/13,x∈(π/2,3π/4),求sinx
∵x∈(π/2,3π/4)
∴x+π/4∈(3π/4,π)
即cos(x+π/4)=-√[1-sin²(x+π/4)]=-12/13
则sinx
=sin[(x+π/4)-π/4]
=sin(x+π/4)cosπ/4-cos(x+π/4)sinπ/4
=5/13*(√2/2)-(-12/13)*(√2/2)
=5√2/26+12√2/26
=17√2/26

sin(x+π/4)=5/13,x∈(π/2,3π/4),
x+π/4∈(3π/4,π)
所以
cos(x+π/4) =-12/13
所以
sinx=sin(x+π/4-π/4)
=sin(x+π/4)cosπ/4-cos(x+π/4)sinπ/4
=5/13 ·√2/2+12/13·√2/2
=17√2/26