已知a为锐角,且tana=根号2-1,函数f(x)=2x.tan2a+sin(2a+派/4),数列{An}的首项a1=1

答：只要求f(x)表达式吗?
锐角a满足：tana=√2-1>0
tan(2a)=2tana /(1-tan²a)
=2(√2-1) /[1-(√2-1)²]
=2(√2-1)/(2√2-2)
=1
所以：2a=π/4,a=π/8
f(x)=2xtan(2a)+sin(2a+π/4)
=2x+sin(π/4+π/4)
=2x+1
所以：f(x)=2x+1
An数列首项A1=1
A(n+1)=f(An)=2An +1
A(n+1)-An=An +1