精品偷拍一区二区三区,亚洲精品永久 码,亚洲综合日韩精品欧美国产,亚洲国产日韩a在线亚洲

  • <center id="usuqs"></center>
  • 
    
  • 求值sin25π/6+cos25π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(-5π/6)×cos(-5π/3)

    求值sin25π/6+cos25π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(-5π/6)×cos(-5π/3)
    數(shù)學人氣:988 ℃時間:2020-09-11 21:24:57
    優(yōu)質(zhì)解答
    sin25π/6+cos25π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(-5π/6)×cos(-5π/3)
    =sin(4π+π/6)+cos(8π+π/3)+tan(-25π/4+6π)+sin(-7π/3+2π)×cos(-13π/6+2π)-sin(π-π/6)×cos(-5π/3+2π)
    =sinπ/6+cosπ/3+tan(-π/4)+sin(-π/3)×cos(-π/6)-sin(π/6)×cos(π/3)
    =sinπ/6+cosπ/3-tanπ/4-sinπ/3×cosπ/6-sin(π/6)×cos(π/3)
    =1/2+1/2-1-√3/2×√3/2-1/2×1/2
    =-3/4-1/4
    =-1
    我來回答
    類似推薦
    請使用1024x768 IE6.0或更高版本瀏覽器瀏覽本站點,以保證最佳閱讀效果。本頁提供作業(yè)小助手,一起搜作業(yè)以及作業(yè)好幫手最新版!
    版權所有 CopyRight © 2012-2024 作業(yè)小助手 All Rights Reserved. 手機版