sina=5⼀13,a属于（π⼀2,π），求sin(a+π⼀4),cos(a-π⼀6），tan(a+π⼀3)

sin(a+π/4)=sinacosπ/4-cosasinπ/4=17√2/26
cos(a-π/6)=cosacosπ/6-sinasinπ/6
tan(a+π/3)=(tana+tanπ/3)/(1-tanatanπ/3)

因为sina=5/13,a属于（π/2,π）
sin^2 a+cos^2 a=1
所以cosa=-12/13
所以tan a=5/13x(-13/12)=-5/12
sin(a+π/4)=sina cos45+sin45 cosa=5/13 x(√2/2)-12/13x( √ 2/2)=-7√ 2/13
cos(a-π/6=cosa cos30+sina sin30=-12/13 x (√3/2)+(5/13)x(1/2)=(-6√3/13)+5/26=(5-12√3)/26
tan(a+π/3)=[-5/12+√3]/(1+5/12 x√3)=(12√3-5)/(12+5√3)=(12√3-5)(12-5√3)/69
=（144√3-180-60+25√3）/69
= (169 √3-240)/69