爱问 爱问共享资料 爱问分类
首页

cos4π相关问答

  • 问: 15+cos16π/15=?

    答:0.5

    答:cos2π=cos4π=cos8π=cos16π=1 所以 cos2π/15+cos4π/15+cos8π/15+cos16π/15=4/15

    学习帮助 2个回答 推荐

  • 问: cosπ9·cos2π9·cos3π9·cos4π9 等于多少?

    答:cosπ/9·cos2π/9·cos3π/9·cos4π/9 =(2sinπ/9cosπ/9·cos2π/9·cosπ/3·cos4π/9)/(2sinπ/9) =(2sin2π/9·cos2π/9·(1/2)·cos4π/9)/(4sinπ/9) =((1/2)2sin4π/9·cos4π/9)...

    数学 1个回答 推荐

  • 问: 数学

    答:cos2π/9*cos4π/9*cos8π/9 =(8sin2π/9*cos2π/9*cos4π/9*cos8π/9)/(8sin2π/9) 同时乘除一个8sin2π/9原式值不变 =(4sin4π/9*cos4π/9*cos8π/9)/(8sin2π/9) 因为2sinacosa=sin2a =(...

    答:由sin2a=2sinacosa 得到 cos2π/9*cos4π/9*cos8π/9 =(8sin2π/9*cos2π/9*cos4π/9*cos8π/9)/(8sin2π/9) =(4sin4π/9*cos4π/9*cos8π/9)/(8sin2π/9) =(2sin8π/9*cos8π/9)/...

    数学 2个回答 推荐

  • 问: 一道数学题

    答:解:16sin(π/17)cos(π/17)cos(2π/17)cos(4π/17)cos(8π/17) =8sin(2π/17)cos(2π/17)cos(4π/17)cos(8π/17) =4sin(4π/17)cos(4π/17)cos(8π/17) =2sin(8π/17)cos(8π/17...

    答:cos(π/17)cos(2π/17)cos(4π/17)cos(8π/17) =[1/2sin(π/17)][2sin(π/17)cos(π/17)]cos(2π/17)cos(4π/17)cos(8π/17) =1/[2sin(π/17)]sin(2π/17)cos(2π/17)cos(4π/1...

    数学 3个回答 推荐

  • 问: 18+cos16π/18=?

    答:cos2π/18+cos4π/18+cos8π/18+cos16π/18= =cos2π/18+cos4π/18+cos8π/18-cos2π/18 =cos4π/18+cos8π/18(化积) =2cos6π/18cos2π/18 =2*(1/2)*cosπ/9 =cosπ/9

    答:首先cos6π/18=cos(π-2π/18)=-cos2π/18 其次cos4π/18=cos(6π/18-2π/18)=cosπ/3cosπ/9-sinπ/3sinπ/9 cos8π/18=cos(6π/18+2π/18)=cosπ/3cosπ/9+sinπ/3sinπ/9 因而cos2π/18...

    学习帮助 2个回答 推荐

  • 问: 证明 cos2π

    答:原式=[(2sin7兀/2)/(2sin2兀/7)]*(cos2兀/7+cos4兀/7+cos6兀/7)=[1/(2sin2兀/7)]*(sin4兀/7+sin6兀/7-sin2兀/7+sin8兀/7-sin4兀/7)=-1/2。

    答:似曾相识 在复数范围内, 方程z^7=1有7个不相等的复数根, 分别是cos2kπ/7+isin2kπ/7(k=0,1,2,3,4,5,6) 这7个根的和等于0, 如果设a=π/7 即[cos0+cos2a+cos4a+cos6a+cos8a+cos10a+cos12a] +i[sin0+sin2a...

    数学 2个回答 推荐

  • 问: 数学二倍角

    答:原式=[sin(2π/7)*cos(2π/7)*cos(4π/7)*cos(6π/7)]/[sin(2π/7)] =[sin(4π/7)*cos(4π/7)*cos96π/7)]/[2sin(2π/7)] =[sin(8π/7)*cos(6π/7)]/[4sin(2π/7)] =[sin(π+π/7...

    学习帮助 1个回答 推荐

  • 问: 三角函数计算题

    答:cos(2π/7)+cos(4π/7)+cos(6π/7) =2cos(3π/7)cos(π/7) + 2cos^(3π/7)-1 =2cos(3π/7)[cos(π/7)+cos(3π/7)]-1 =4cos(3π/7)cos(2π/7)cos(π/7)-1 =4cos(3π/7)cos(2π/7...

    学习帮助 1个回答 推荐

  • 问: 一道[三角恒等变换]的化简题

    答:用积化和差公式轻松解决(pi表示 派): 原式=1/2(cos(-2pi/5)+cos(pi))+1/2(cos(-3pi/5)+cos(pi)) =-1+1/2(cos(2pi/5)-cos(2pi/5)) =-1

    数学 1个回答

热点检索
  • 1-20
  • 21-40
  • 41-60
  • 61-80
  • 81-100
  • 101-120
  • 121-140
  • 141-160
  • 161-180
返回
顶部