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为什么[sin(n+1/2)*x]/sin(x/2)有界
题目详情
为什么[sin(n+1/2)*x]/sin(x/2)有界
▼优质解答
答案和解析
n=0 =>sin(n+1/2)x/sin(x/2)=1/2 有界n>=1 => sin(n+1/2)x/sin(x/2)=sinnxcos(x/2)/[sin(x/2)]+cosnxsin(n+1/2)xcos(x/2)/[sin(x/2)cos(x/2)]=[sin(n+1)x+sinnx]/sinx−n≤sin(nx)/sin(x)≤n−n-1≤sin(n+...
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