sinx+sin3x+sin5x+......sin(2n-1)⼀cosx+cos3x+cos5x+......cos(2n-1)= sin2x+sin4x+......sin 2nx ⼀cos2x

解:

(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<7且n∈N*

p<1时数列单调减:

(9/10)(n+3)/(n+2)<1

(n+3)/(n+2)<10/9

1/(n+2)<1/9

解得n>7且n∈N*

当n=7时取到最大值，此时有:

p=a(n+1)/a(n)=a(8)/a(7)=1

故n=7或n=8时该数列取到最大值

最大值为:

a(8)=a(7)=9×(9/10)^7