Question

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.)