1.化简:sin(4k-1/4π -a)+cos(4k+1/4π -a)(k∈Z)

问题描述:

1.化简:sin(4k-1/4π -a)+cos(4k+1/4π -a)(k∈Z)

原式=sin(kπ-π/4-a)+cos(kπ+π/4-a)
k是偶数
则原式=sin(-a-π/4)+cos(π/4-a)
=-sinacosπ/4-cosasinπ/4+cosπ/4cosa+sinπ/4sina
=0
k是奇数
则原式=sin(a-π/4)+cos(π/4+a)
=sinacosπ/4-cosasinπ/4+cosπ/4cosa-sinπ/4sina
=0
所以原式=0