[1/cosacos(a+b)]+[1/cos(a+b)cos(a+2b)]+....+{1/COS[a+(n-1)b]COS(a+nb)}
热心网友
Sn=[1/cosacos(a+b)]+[1/cos(a+b)cos(a+2b)]+。。。。+{1/COS[a+(n-1)b]COS(a+nb)}1)b=mπ,Sn=[n(-1)^m]/cosa^2。2)b≠mπ有sinb/cos(a+kb)cos(a+(k+1)b)]=sin[a+(k+1)b)-(a+kb)]/cos(a+kb)cos(a+(k+1)b)]=[cos(a+kb)sin(a+(k+1)b)-sin(a+kb)cos(a+(k+1)b)]/[cos(a+kb)cos(a+(k+1)b)]==[tg(a+(k+1)b)-tg(a+(kb)],所以，sinbSn=[tg(a+2b)-tg(a+b)]+。。+[tg(a+(n+1)b)-tg(a+(nb)]==tg(a+(n+1)b)-tg(a+b),Sn=[tg(a+(n+1)b)-tg(a+b)]/sinb。补：cos（a+kπ）=cosacoskπ-sinasinkπ=（-1）^kcosa。cos（a+lmπ）cos（a+(l+1)mπ）=（-1）^[(2l-1)m]cosa^2。=（-1）^mcosa^2。所以b=mπ,Sn=[n(-1)^m]/cosa^2。。