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
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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