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