已知sin(a+π/6)=12/13,a+π/6属于第二象限，求cosa与sina的值
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sin(a+π/6)=12/13
a+π/6∈(π/2,π)
cos(a+π/6)=-5/13
cosa
=cos(a+π/6-π/6)
=cos(a+π/6)cosπ/6+sin(a+π/6)sinπ/6
=-5/13*√3/2+12/13*1/2
=(12-5√3)/26
sina
=sin(a+π/6-π/6)
=sin(a+π/6)cosπ/6-cos(a+π/6)sinπ/6
=12/13*√3/2+5/13*1/2
=(12√3+5)/26
a+π/6∈(π/2,π)
cos(a+π/6)=-5/13
cosa
=cos(a+π/6-π/6)
=cos(a+π/6)cosπ/6+sin(a+π/6)sinπ/6
=-5/13*√3/2+12/13*1/2
=(12-5√3)/26
sina
=sin(a+π/6-π/6)
=sin(a+π/6)cosπ/6-cos(a+π/6)sinπ/6
=12/13*√3/2+5/13*1/2
=(12√3+5)/26
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