∵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),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