f(π/9)=sin(3π+π/4）=sin(π+π/4）=-sinπ/4=-√2/2
f（a/3+π/4）=1/2=cos(π/3）
a/3+π/4=π/3
a/3=π/12
a=π/4
则cos2a=0不应该是f(π/9)=sin(π/3π+π/4）吗？？f(π/9)=sin(π/3+π/4）=sin(7π/12）=sin（π-5π/12）=sin5π/12f（a/3+π/4）=1/2=cos(π/3）a/3+π/4=π/3a/3=π/12a=π/4则cos2a=0【？】