cos(π/3)的这个连根式展开公式怎么证明？ cos7π诱导公式怎么求

cos(π/3)的这个连根式展开公式怎么证明？cos7π诱导公式怎么求

sin π/3 =根号3/2，怎么来的？怎么解？请给出详细过程，谢谢

楼主你好，直角三角形满足勾股定理，设直角三角形勾为1,所以股为根号3，弦为2，sin60是正弦，对边比斜边，在本题中就是股比弦，也就是根号3比2,就是你说的根号3/2，回答完毕，希望楼主采纳，谢谢

1.cosπ/3=1/2

2.cosπ/5cos2π/5

=2sin(π/5)*cos(π/5)*cos(2π/5) / 2sin(π/5)

=sin(2π/5)*cos(2π/5) / 2sin(π/5)

=sin(4π/5) / 4sin(π/5)

=sin(π/5) / 4sin(π/5)

=1/4

3.cos(π/7)cos(2π/7)cos(3π/7)

=2sin(π/7)cos(π/7)cos(2π/7)cos(3π/7) / 2sin(π/7)

=sin(2π/7)cos(2π/7)(cos3π/7) / 2sin(π/7)

=sin(4π/7)cos(3π/7) / 4sin(π/7)

=sin(8π/7) / 8sin(π/7)

=sin(π/7) / 8sin(π/7)

=1/8

一般结论：

cos[π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)]=1/(2^n)

证明：

cos[π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)]

=2sin[π/(2n-1)]cos[π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)] / 2sin[π/(2n-1)]

【分子分母补2sin[π/(2n-1)]】

=sin[2π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)] / 2sin[π/(2n-1)]

=2sin[2π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)] / 4sin[π/(2n-1)]

=sin[4π/(2n-1)]……cos[nπ/(2n-1)] / 4sin[π/(2n-1)]

=……

=2sin[nπ/(2n-1)]cos[nπ/(2n-1)] / (2^n)sin[π/(2n-1)]

=cos[2nπ/(2n-1)] / (2^n)sin[π/(2n-1)]

=sin[π/(2n-1)] / (2^n)sin[π/(2n-1)] 【cos[2nπ/(2n-1)] =sin[π/(2n-1)] 】

=1/(2^n)

重点：不断地利用二倍角公式：sin2θ=2sinθcosθ

希望采纳~~~~~~~~

照你的意思，(1+cosx/2)^a`=a(1+cosx/2)^(a-1)

不过你可以设t=cosx/2再用你的方法计算，之后记得用x代入t就行

1、cos(a+π/2)=-sina=(2√6)/5；

2、sin[π/2-(π/6+a)]=cos(π/3-a)=√3/3