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

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

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