x∈（0,π/2）x+π/6∈（π/6,2π/3）所以sin(x+π/6)>0sin²(x+π/6)+cos²(x+π/6)=1所以sin(x+π/6)=12/13cosa=cos[(x+π/6)-π/6]=cos(x+π/6)cosπ/6+sin(x+π/6)sinπ/6=(5√3+12)/26