Question

求证：【cos(kπ-α)cos(kπ+α)】/{sin【(k+1)π+α】cos【(k+1)π+α】}=-1

求证：【cos(kπ-α)cos(kπ+α)】/{sin【(k+1)π+α】cos【(k+1)π+α】}=-1

Answer 1

k是整数吗？
好像证不出来，很有可能做错了。检查一下题目，
= = = = = = = = =

证明：(1)当 k是奇数时,
cos(kπ-α)= -cos(-α)= -cos(α),
cos(kπ+α)= -cos(α),
sin[(k+1)π+α]=sin(α),
cos[(k+1)π+α]=cos(α).
所以 [cos(kπ-α)cos(kπ+α)]/{sin[(k+1)π+α]cos[(k+1)π+α]}
=[-cos(α)]*[-cos(α)]/[sin(α)cos(α)]
= cot(α).
(这一步有问题！！！)
(2)当 当 k是偶数时,
cos(kπ-α)=cos(-α)=cos(α),
cos(kπ+α)=cos(α),
sin[(k+1)π+α]= -sin(α),
cos[(k+1)π+α]= -cos(α).
所以 [cos(kπ-α)cos(kπ+α)]/{sin[(k+1)π+α]cos[(k+1)π+α]}
=[cos(α)*cos(α)]/{[-sin(α)][-cos(α)]}
= cot(α).

综上,
[cos(kπ-α)cos(kπ+α)]/{sin[(k+1)π+α]cos[(k+1)π+α]}=cot(α),
(k为任意整数.)

Answer 2

积化和差