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.
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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

sinx+sin5x+sin3x=0 x->1时(sin5x-sin3x)/sinx的极限是多少 函数sin3x/sinx-sin5x/sinx x∈(0,π/2)求值域我要详细的过程谢谢 由此证明sinx-sin3x+sin5x=sin6x/2cosx,cosx≠0cosx+cos3x+cos5x +...+ cos(2n-1)x= sin2nx/2sinx由此证明sinx - sin3x + sin5x = sin6x/2cosx ,cosx≠0 lim(x趋于0)sin5x-sin3x/sinx=可以直接用等价无穷小来做吗?分子加减是不是不能用? sin4x+sin5x+sin6x=0,sinx+sin2x+sin3x=0 找出所有解,有明白的麻烦详细点, limx→0sin3x/sin5x,求极限 lim（x→0）sin5X/sin3X lim(x-0)(sin3x*sin5x)/(x^2) 求极限limx-0 sin3x/sin5x 三角函数 1.解x,2（cosx）平方+3sinx=3 2.证明（sinx-sin2x+sin3x）/(cosx-cos2x+cos3x)=tan2x3.证明sin5x-sin3x+sinx=o sinx+sin2x+sin3x 求最值 sinx*sin2x*sin3x=? sin3x/sinx+cos3x/cosx 求极限,lim(x→π)sin3x/sin5x 计算lim（x→0）sin3x/sin5x 函数F(x)=Sin3X+Sin5X的最小正周期 lim(x→0+)[ln(sin5x)]/[ln(sin3x)] 这题咋做