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问题：1.化简:sin(4k-1/4π-a)+cos(4k+1/4π-a)(k∈Z)
答案：↓↓↓ 李建清的回答： 网友采纳 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 | 
