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>
  • 
    
  • cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×cos(6pai/15)×cos(7pai/15)求值

    cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×cos(6pai/15)×cos(7pai/15)求值
    數(shù)學(xué)人氣:507 ℃時(shí)間:2020-07-10 02:38:30
    優(yōu)質(zhì)解答
    原式=[1/(sinπ/15)]•[sin(π/15)cos(π/15)cos(2π/15)..cos(7π/15)]
    =[sin(8π/15)/8sin(π/15)]•[cos(3π/15)cos(5π/15)cos(6π/15)cos(7π/15)]
    =...
    =[sin(4π/5)/2^7sin(π/5)]
    =1/128
    字?jǐn)?shù)限制
    我來回答
    類似推薦
    請(qǐng)使用1024x768 IE6.0或更高版本瀏覽器瀏覽本站點(diǎn),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(quán)所有 CopyRight © 2012-2024 作業(yè)小助手 All Rights Reserved. 手機(jī)版