请按定义证明数列发散 {sinn} 我实在不明白如何证明数列发散？！谢谢各位 证明数列an=sinn发散详细过程？
\u5982\u4f55\u8bc1\u660e\u6570\u5217\uff5bsin n\uff5d\u4e3a\u53d1\u6563\u6570\u5217\uff1f\u770b\u8fd9\u4e2a\uff0chttp://www.doc88.com/p-309514163999.html
\u7b2c\u4e00\u9053\u9898\uff1a\u5982\u679c\u8bc1\u660e\u6570\u5217{sinn}\u6781\u9650\u4e0d\u5b58\u5728\u3002
\u6839\u636e\u53d1\u6563\u5b9a\u4e49\uff0c\u5219\u8bc1\u660e{sinn}\u4e0d\u6536\u655b\uff0c\u662f\u53d1\u6563\u7684\u3002
\u9996\u5148\u8981\u5f15\u5165
\u5b9a\u4e49\uff1a\u5b50\u5217\uff1a
\u5b9a\u7406\uff1a\u6570\u5217lim An = a\uff0c\u7b49\u4ef7\u4e8e An\u4e2d\u4efb\u610f\u5b50\u5217\u7684\u6781\u9650\u4e5f\u662fa
\u8bc1\u660e\uff1a\u53cd\u8bc1\u6cd5\uff1a\u5047\u8bbesin n\u6709\u6781\u9650\uff1alim sinn = a
\u7531\u4e09\u89d2\u51fd\u6570\u516c\u5f0f\u53ef\u77e5\uff1a|sinn|<=1\uff0c\u5373sinn\u662f\u6709\u754c\u7684
\u7531\u4e09\u89d2\u51fd\u6570\u516c\u5f0f\uff1asin(a+b) = sina.cosb + cosa.sinb
\u53ef\u77e5\u7b49\u5f0f\u6210\u7acb\uff1asin(n+1) - sin(n-1) = 2sin1cosn
\u5bf9\u7b49\u5f0f\u4e24\u8fb9\u53d6\u6781\u9650\uff1a
\u7b49\u5f0f\u5de6\u8fb9\u5206\u6790\uff1a
\u7531\u5b50\u5217\u5b9a\u4e49\u53ef\u77e5\uff1asin(n+1)\u3001sin(n-1)\u90fd\u662fsinn\u7684\u5b50\u5217\uff0c
\u7531\u5b50\u5217\u5b9a\u7406\u53ef\u77e5\uff1alim sin(n+1) = lim sin(n-1) = lim sinn = a
\u7b49\u5f0f\u5de6\u8fb9\uff1a lim sin(n+1) - lim sin(n-1) = 0
\u7b49\u5f0f\u53f3\u8fb9\u5206\u6790\uff1a2sin1\u662f\u4e00\u4e2a\u5e38\u6570\uff0clim 2sin1 = 2sin1
\u7efc\u5408\u5206\u6790\u540e\uff1a lim sin(n+1) - lim sin(n-1) = 0= 2sin1.lim cosn
lim cosn = 0
\u7531\u4e09\u89d2\u51fd\u6570\u516c\u5f0f\uff1asin(a+b) = sina.cosb + cosa.sinb
\u53ef\u77e5\u7b49\u5f0f\u6210\u7acb\uff1asin2n = sin(n+n) = 2sinn.cosn
\u5bf9\u7b49\u5f0f\u4e24\u8fb9\u53d6\u6781\u9650\uff1alim sin2n = lim 2sinn.cosn
\u7b49\u5f0f\u5de6\u8fb9\u5206\u6790\uff1a
\u7531\u5b50\u5217\u5b9a\u4e49\u53ef\u77e5\uff1asin2n\u90fd\u662fsinn\u7684\u5b50\u5217\uff0c
\u7531\u5b50\u5217\u5b9a\u7406\u53ef\u77e5\uff1alim sin2n = lim sinn = a
\u7b49\u5f0f\u53f3\u8fb9\u5206\u6790\uff1a\u7531\u4e0a\u9762lim cosn = 0\uff0c\u53ef\u5f97\uff1a\u53f3\u8fb9\u6781\u9650\u4e3a0
\u7efc\u5408\u5206\u6790\u540e\uff1alim sin2n = a = lim 2sinn.cosn = 0\uff0c\u53ef\u5f97a = 0
\u7531\u4e09\u89d2\u51fd\u6570\u516c\u5f0f\uff1asin^2 a +cos^2 b = 1
\u7b49\u5f0f\u4e24\u8fb9\u53d6\u6781\u9650\uff1alim sin^2 a + lim cos^2 b = lim 1 = 1
a^2 + a^2 = 1
0^2 + 0^2 = 1
0 = 1 \u77db\u76fe\uff0c\u5047\u8bbe\u4e0d\u6210\u7acb
而sin(n+2)-sinn=2cos(n+1)sin1，得lim2cos(n+1)sin1=a-a=0，则limcos(n+1)=0,limcosn=0。
则a=limsinn=lim√(1-cos^2 n)=1。
又 sin2n=2sinncosn，两边取极限，得a=2a×0，矛盾。
所以数列sin n是发散的。
扩展阅读：扫一扫题目出答案 ... sinπx诱导公式 ... 证明数列an 1 n发散 ... 证明sin 2n+1 发散 ... 证明数列sinn发散 ... 如何证明数列收敛题目 ... sin n分之一 收敛还是发散 ... 怎么证明∑sin nx发散 ... 证明sinn发散 n为整数 ...
