sinx+sin3x+sin5x+.sin(2n-1)/cosx+cos3x+cos5x+.cos(2n-1)= sin2x+sin4x+.sin 2nx /cos2xsinx+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

Question

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

Answer 1

(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