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求下列三角函数值:sin7π/2+cos2π/3+tan(-25?

2011-03-30 21:12:41j***
求下列三角函数值:sin7π/2+cos2π/3+tan(-25π/4) 分析下,谢谢求下列三角函数值:sin7π/2+cos2π/3+tan(-25π/4)分析下,谢谢:原式=sin[4π-(π/2)]+cos(2π/3)-tan[6π+(π/?

最佳回答

  • 原式=sin[4π-(π/2)]+cos(2π/3)-tan[6π+(π/4)] =sin[-(π/2)]+cos(2π/3)-tan(π/4) =-1-(1/2)-1 =-5/2
    2011-03-30 21:28:27
  • 很赞哦! (48)