www.whkt.net > Cosx/2乘Cosx/4一直乘到Cosx/2^n 等于多少? 大神求解!!

Cosx/2乘Cosx/4一直乘到Cosx/2^n 等于多少? 大神求解!!

解:原式=[2^nsin(x/2^n) *cosx/2cosx/4..cosx/2^n]/2^nsin(x/2^n)=sinx/[2^nsin(x/2^n)]

cosx/2*cosx/4…*cosx/2的n次方=cosx/2*cosx/4…*cosx/2的n次方 *2的n次方*sinx/2的n次方/[2的n次方*sinx/2的n次方]=cosx/2*cosx/4…*cosx/2的n-1次方 *2的n-1次方*sinx/2的n-1次方/[2的n次方*sinx/2的n次方]=..=2cosx/2sinx/2/[2的n次方*sinx/2的n次方]=sinx/[2的n次方*sinx/2的n次方]

两边乘SIN(X/2^N)则有SIN(X/2^N)*COS(X/2^N)=(1/2)SIN(X/2^(N-1))所以原式=(1/2^N)SINX

因cos x /2cosx/4…cosx/2^n=[cosx/2*cosx/4*.*2sinx/2^n*cosx/2^n]/(2sinx/2^n)=[cosx/2*cosx/4**sinx/2^(n-1)]/(2sinx/2^n)=(cosx/2sinx/2)/[2^(n-1)*sin(x/2^n]=sinx/[2^n*sin(x/2^n)] 所以lim (n趋近正无穷) cos x /2cosx/4…cosx/2^n=lim (n趋近正无穷) sinx/[x*sin(x/2^n)/(x/2^n)]=(sinx)/x

cosx/2 cosx/4 ……cosx/(2^n) =[cosx/2 cosx/4 ……cosx/(2^n) sinx/(2^n) ]/sinx/(2^n) =[cosx/2 cosx/4 ……cosx/(2^n-1) sinx/(2^n-1) ]/2sin/sinx/(2^n) =sinx/2^nsinx/(2^n) =sinx/sinx/(2^n)/(1/2^n) 当n趋向于无穷大时,sinx/(2^n)/(1/2^n)趋向于1, 所以lim cosx/2 cosx/4 ……cosx/(2^n)=sinx

1、本题是涉及余弦的连乘问题;2、解答方法是: A、反复使用正弦二倍角公式 B、运用重要极限 sinx/x = 1.3、详细解答如下:

分子分母同乘sin(x/2^n)分子一步步 可等于(sinx)/2^n分母 = sin(x/2^n)因为 lim x/sinx = 1 x趋于0时所以 lim (x/2^n)/sin(x/2^n) = 1而分子 = sinx/x * x/2^n所以原题极限 = sinx/x

解法如下:y=cos(x/2)cos(x/4).cos(x/2^n) 两边x2sin(x/2^n)2ysin(x/2^n)=cos(x/2)cos(x/4).cos(x/2^n)2sin(x/2^n)2ysin(x/2^n)=cos(x/2)cos(x/4).sin(x/2^(n-1)) x2:2ysin(x/2^n)=cos(x/2)cos(x/4).sin(x/2^(n-2)).2^kysin(x/2^n)=cos(x/2)cos(x/4).sin(

解:cosx乘以2cosx=2cos^2x=cos2x+1f(x)=m*n=(1,2cosx)(√3sin2x+1,cosx)=√3sin2x+1+2cos^2=√3sin2x+1+cos2x+1=2sin(2x+π/6)+2所以最小正周期为2π/2=π.解答完毕.

求函数列fn(x)= cosx/2*cosx/4*cosx/8cosx/2^n当n趋于∞时的极限当x不等于k*pi时,对任意fn,乘以sin(x/2^n)/sin(x/2^n),利用sin(2x)=2sin(x)*cos(x)得fn(x)=sin(x)/(2^n*sin(x/2^n)),极限为sin(x),易验证x=k*pi时亦符合以为极限sin(x).兄弟,我知道帐号缺财富值了,帮帮忙……

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