(1)化简.(2)求函数y=2-sin2x+cosx的最大值及相应的x的值.【考点】;.【专题】计算题;转化思想.【分析】(1)由诱导公式对所给的解析式化简,即可得到结果(2)将函数y=2-sin2x+cosx变为关于cosx的二次函数,进行配方,再根据余弦函数的有界性判断出最值以及相应的x的值【解答】解:(1)原式(2)y=2-sin2x+cosx=cos2x+cosx+1=(cosx+)2+当cosx=1时,函数取得最大值为3,此时x=2kπ,k∈z【点评】本题考查运用诱导公式化简求值以及求三角函数的最值,解题的关键是熟练掌握公式并能用之进行变化化简.声明:本试题解析著作权属菁优网所有,未经书面同意,不得复制发布。答题: 难度:0.74真题:3组卷:4
解析质量好中差求sin(2nx+4π/3)+cos(nπ+2π/3)(n∈Z)的值
求sin(2nx+4π/3)+cos(nπ+2π/3)(n∈Z)的值 5
2(sin(nx+2π/3)+1)cos(nπ+2π/3)
详细的过程
=sin2(nx+2π/3)+cos(nπ+2π/3)
(n∈Z)=2sin(nx+2π/3)cos(nπ+2π/3)+cos(nπ+2π/3)
(n∈Z)=2(sin(nx+2π/3)+1)cos(nπ+2π/3)
其他回答 (1)
当N为偶数时答案是负根号3.当n为奇数是答案为0
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