sinx+sin3x+sin5x+.sin(2n-1)/cosx+cos3x+cos5x+.cos(2n-1)= sin2x+sin4x+.sin 2nx /cos2x+cos4x+.cos2nx =

题目

sinx+sin3x+sin5x+.sin(2n-1)/cosx+cos3x+cos5x+.cos(2n-1)= sin2x+sin4x+.sin 2nx /cos2x+cos4x+.cos2nx = 已知数列an通项公式（n+2）*（9/10）的n次方,求n为何值an最大,并求最大值.

答案

(1) 设S=sinx+sin3x+sin5x+.+sin(2n-1)x S=sin(2n-1)x+sin(2n-3)x+.+sinx 上下对应项相加得(和差化积):2S=2sin(nx)cos(n-1)x+2sin(nx)cos(n-3)x+.+2sin(nx)cos(n-3)x+2sin(nx)cos(n-1)x S=sin(nx)[cos(n-1)x+cos(n-3...