求cosπ/9*cos2π/9*cos3π/9*cos4π/9
答案:2 悬赏:70 手机版
解决时间 2021-02-01 08:37
- 提问者网友:蓝莓格格巫
- 2021-02-01 00:50
求cosπ/9*cos2π/9*cos3π/9*cos4π/9
最佳答案
- 五星知识达人网友:底特律间谍
- 2021-02-01 02:17
cosπ/9*cos2π/9*cos3π/9*cos4π/9=(cosπ/9*cos2π/9*cos3π/9*cos4π/9)*8(sinπ/9)/8(sinπ/9)∵2cosπ/9*sinπ/9=sin2π/9∴原式=(4sin2π/9*cos2π/9*cos3π/9*cos4π/9)/8(sinπ/9)∵2sin2π/9*cos2π/9=sin4π/9∴原式=(2sin4π/9*cos3π/9*cos4π/9)/8(sinπ/9)=(cos3π/9*sin8π/9)/8(sinπ/9)∵ sinπ/9=sin8π/9……诱导公式 ∴原式=(cosπ/3*sin8π/9)/8(sinπ/9)=(cosπ/3)/8=1/16======以下答案可供参考======供参考答案1:x=cosπ/9*cos2π/9*cos3π/9*cos4π/9y=sinπ/9*sin2π/9*sin3π/9*sin4π/9则xy=[1/(2^4)]sin2π/9*sin4π/9*sin6π/9*sin8π/9设z=sin2π/9*sin4π/9*sin6π/9*sin8π/9因为sinπ/9*sin3π/9*sin5π/9*sin7π/9=sin8π/9*sin10π/9*sin12π/9*sin14π/9所以y=z所以x=1/(2^4)~~~通常~~~cosπ/(2n+1)*cos2π/(2n+1)*cos3π/(2n+1)*...*cosnπ/(2n+1)=1/(2^n),n是正整数所以答案= 1/(2^4) = 1/16供参考答案2:cosπ/9*cos2π/9*cos3π/9*cos4π/9=(sinπ/9cosπ/9*cos2π/9*cos3π/9*cos4π/9) /(sinπ/9)=1/2(sin2π/9*cos2π/9*cos3π/9*cos4π/9) /(sinπ/9)=1/4(sin4π/9*cos3π/9*cos4π/9) /(sinπ/9)=1/8(sin8π/9*cos3π/9) /(sinπ/9)=1/8(sinπ/9*cos3π/9) /(sinπ/9)=1/8*cos3π/9=1/8*1/2=1/16供参考答案3:cosπ/9*cos2π/9*cos3π/9*cos4π/9=sinπ/9*cosπ/9*cos2π/9*cos3π/9*cos4π/9/sinπ/9=1/2sin2π/9cos2π/9*1/2*cos4π/9/sinπ/9=1/2*1/2*1/2*sin4π/9*cos4π/9/sinπ/9=1/2*1/2*1/2*1/2*sin8π/9/sinπ/9又sin8π/9=sinπ/9故原式=1/16
全部回答
- 1楼网友:低音帝王
- 2021-02-01 03:09
这下我知道了
我要举报
如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息,可以点下面链接进行举报!
大家都在看
推荐资讯