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.
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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.
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.
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.
(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)x+.+cos(n-3)x+cos(n-1)x]-------------------[1] 设s=cosx+cos3x+cos5x+.+cos(2n-1)x s=cos(2n-1)x+cos(2n-3)x+.+cosx 上下对应项相加得(和差化积):2s=2cos(nx)cos(n-1)x+2cos(nx)cos(n-3)x+.+2cos(nx)cos(n-3)x+2cos(nx)cos(n-1)x s=cos(nx)[cos(n-1)x+cos(n-3)x+.+cos(n-3)x+cos(n-1)x]------------------[2] [1]/[2]得:原式=S/s=sin(nx)/cos(nx)=tan(nx)(n∈N*) (2) 和(1)的解法类似:设S'=sin2x+sin4x+.+sin(2nx) S'=sin(2nx)+sin(2n-2)x+.+sin2x 上下对应项相加得(和差化积):2S'=2sin(n+1)xcos(n-1)x+2sin(n+1)xcos(n-3)x+.+2sin(n+1)xcos(n-3)x+2sin(n+1)xcos(n-1)x S'=sin(n+1)x[cos(n-1)x+cos(n-3)x+.+cos(n-3)x+cos(n-1)x]---------------[3] 设s'=cos2x+cos4x+.+cos(2nx) s'=cos2nx+cos(2n-2)x+.+cos2x 上下对应项相加得(和差化积):2s'=2cos(n+1)xcos(n-1)x+2cos(n+1)xcos(n-3)x+.+2cos(n+1)xcos(n-3)x+2cos(n+1)xcos(n-1)x s'=cos(n+1)x[cos(n-1)x+cos(n-3)x+.+cos(n-3)x+cos(n-1)x]---------------[4] [3]/[4]得:原式=S/s=sin(n+1)x/cos(n+1)x=tan(n+1)x(n∈N*) (3) a(n)=(n+2)×(9/10)^n>0,n∈N* 设p=a(n+1)/a(n)=[(n+3)×(9/10)^(n+1)]/[(n+2)×(9/10)^n] =(9/10)(n+3)/(n+2) p>1时数列单调增:(9/10)(n+3)/(n+2)>1 (n+3)/(n+2)>10/9 1/(n+2)>1/9 解得n