三角函数Sn=sinπ/12（sinπ/）6+sin2π/6+……+sinnπ/6）由三角积差公式怎么得到Sn=1/2（cosπ/12-cos（n+1/2）π/6）

Question

三角函数
Sn=sinπ/12（sinπ/）6+sin2π/6+……+sinnπ/6）
由三角积差公式怎么得到
Sn=1/2（cosπ/12-cos（n+1/2）π/6）

Answer 1

答：
积差公式：sinasinb=(1/2)[cos(a-b)-cos(a+b)]
Sn=sin(π/12)*[sin(π/6)+sin(2π/6)+...+sin(nπ/6) ]
Sn=sin(π/12)*sin(π/6)+sin(π/12)*sin(2π/6)+...+sin(π/12)*sin(nπ/6)
2Sn=[cos(π/6-π/12)-cos(π/6+π/12)]+[cos(2π/6-π/12)-cos(2π/6+π/12)]+.+
[cos(nπ/6-π/12)-cos(nπ/6+π/12)]
2Sn=cos(π/12)-cos(π/4)+cos(π/4)-cos(5π/12)+.-cos(nπ/6+π/12)
2Sn=cos(π/12)-cos(nπ/6+π/12)
（中间各项正负抵消）
所以：
Sn=(1/2)*[cos(π/12)-cos(nπ/6+π/12)]