COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/06 00:29:03
COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程COS(π/17)COS(2π/
COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程
COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程
COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程
答：
cos(π/17)cos(2π/17)cos(4π/17)cos(8π/17)
=[16sin(π/17)cos(π/17)cos(2π/17)cos(4π/17)cos(8π/17)]/16sin(π/17)
=[8sin(2π/17)cos(2π/17)cos(4π/17)cos(8π/17)]/16sin(π/17)
=[4sin(4π/17)cos(4π/17)cos(8π/17)]/16sin(π/17)
=[2sin(8π/17)cos(8π/17)]/16sin(π/17)
=sin(16π/17)/16sin(π/17)
=1/16
cos(π/17)*cos(2π/17)*cos(4π/17)*cos(8π/17)
COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程
倍角公式与和角公式化简或求值cos π/17 * cos 2π/17 * cos 4π/17 * cos 8π/17
求值Cos pai/17 *cos 2pai/17*cos 4pai/17*cos 8pai/17
计算：cos（π/5)+cos（2π/5)+cos(3π/5)+cos(4π/5)
cos^2α+cos^2(α+π/3)-cosαcos(α+π/3)化简
求值:cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11)
求值：cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11)
cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11)
① 利用公式C（a－β)证明,cos(2π-a)=cos a
已知三角型abc三内角a b c 成等差数列,且a-c=π/3,求cos^2a+cos^2b+cos^c的值
已知sina=15/17,cosβ= - 5/13,a∈（π/2,π）,求cos（a+β）,cos（a-β）的值.急用
cos（－17/3π）＝?
cos(-17/6π）=
在△ABC中,根号3cos(B+C)求根号3cos(B+C)+COS(π/2+A）的取值范围
cos(θ-π/2)等于cos(π/2-θ)
2cos(π/2+a)cos(π-a)化简
计算cos(π/5)cos(2π/5)的值