sin3x除以sin5x

籍光雷3183求极限,lim(x→π)sin3x/sin5x -
封食勤15663975237 ______[答案] 当x→π时,sin3x→0且sin5x→0 即是我们所说的0/0型 我们可以用洛必达法则上下分别对x求导,得 lim(x→π)sin3x/sin5x= lim(x→π)3cos3x/5cos5x= lim(x→π)3cos(3π)/5cos(5π)=(-3)/(-5)=3/5 如果有什么疑问可以问我!

籍光雷3183limx→0sin3x/sin5x,求极限 -
封食勤15663975237 ______[答案] lim sin3x/sin5x =lim 3x/(5x) =3/5 ======== 当x趋于0时,sin3x等价于3x,sin5x等价于5x

籍光雷3183高一三角 sinx+sin5x+sin3x=0 -
封食勤15663975237 ______ sin(3x-2x)+sin(3x+2x)+sin3x → sin3xcos2x-cos3xsin2x+sin3xcos2x+cos3xsin2x+sin3x → sin3xcos2x+sin3xcos2x+sin3x → 2sin3xcos2x+sin3x → 2cos2x(sin3x) ∴2cos2x=0或sin3x=0

籍光雷3183limx趋近于0时, sin3x/sin5x=?? -
封食勤15663975237 ______ limx趋近于0时, sin3x/sin5x=3x/5x=3/5(等价无穷小代换)

籍光雷3183函数sin3x/sinx - sin5x/sinx x∈(0,π/2)求值域 -
封食勤15663975237 ______ 解:sin(3x)/sinx-sin(5x)/sinx=[sin(3x)-sin(5x)]/sinx={2cos[(3x+5x)/2]sin[(3x-5x)/2]}/sinx=[2cos(4x)sin(-x)]/sinx=-2cos(4x) 因为:x∈(0,π/2) 所以:4x∈(0,2π) 故:-2cos(4x)∈(-2,2) 因此所求值域为:(-2,2).

籍光雷3183自学高数者的一个白痴问题 急记急记急(sin3x)/(sin5x)的极限 .当x等零时. -
封食勤15663975237 ______[答案] 等价无穷小的替换.sin3x~3x,sin5x~5x,所以就是3x/5x的极限,结果为3/5.

籍光雷3183求极限limx - 0 sin3x/sin5x -
封食勤15663975237 ______[答案] limx-0 sin3x/sin5x=limx-0 [(sin3x/3x)*3x]/[(sin5x/5x)/5x]=limx-0 [(sin3x/3x)*3x]/[(sin5x/5x)/5x]=3/5limx-0[ (sin3x/3x)/(sin5x/5x)]=3/5[limx-0 (sin3x/3x)]/[limx-0 (sin5x/5x)]=3/5*1/2=3/5希望帮你解决了本题.

籍光雷3183limsin3x - ----=?sin5x -
封食勤15663975237 ______[答案] x→0时 sin3x等价于3x,sin5x等价于5x,从而 lim_{x→0}sin3x/sin5x=lim_{x→0}3x/5x=3/5.

籍光雷3183sin3x=sin5x 求x sin3x=sin5x 求x -
封食勤15663975237 ______[答案] sin3x=sin5x 2kπ+3x=5x 或 3x=(2k+1)π-5x => x=kπ或 x=((2k+1)π)/8 k∈Z

籍光雷3183求(sin3x)(sin5x)的原函数 -
封食勤15663975237 ______[答案] cos(a+b)=cosa cosb- sinasinbcos(a-b)=cosa cosb+ sinasinb第二式减第一式,得2sinasinb=cos(a-b)-cos(a+b)即sinasinb=[cos(a-b)-cos(a+b)]/2故(sin3x)(sin5x)=(cos2x-cos8x)/2原函数为(sin2x)/4-(sin8x)/16...