Question

cosπ/15 cos2π/15 cos3π/15 cos4π/15 cos5π/15 cos6π/15 cos7π/15

他们都相乘 求等于多少

Answer 1

像这种题目一般都是有一个通法就是凑用到一个公式：利用二倍角公式sin2x=2sinxcosx

原式=【2sinπ/15cosπ/15 cos2π/15 cos3π/15 cos4π/15 cos5π/15 cos6π/15cos7π/15】/2sinπ/15
=【sin2π/15cos2π/15 cos3π/15 cos4π/15 cos5π/15 cos6π/15 cos7π/15】/2sinπ/15
=【2sin2π/15cos2π/15 cos3π/15 cos4π/15 cos5π/15 cos6π/15 cos7π/15】/4sinπ/15
.
.
=【sin14π/15】/128sinπ/15
因为sin14π/15=sinπ/15
所以原式=【sin14π/15】/128sinπ/15=【sinπ/15】/128sinπ/15=1/128
只加了sinπ/15 因子就可化简其余补2化简就可以了
高考就是用这种通法解题，不要想得太复杂，他不会考你死死的化简，一定有非常简便的方法，其他方法你是不容易想到，比如1楼和2楼的，你是很难找到他们所说化简因子的，而且题目一变你又觉得难了，这种思维你就好好领会它的好处吧！

Answer 2

1/128 用这个式子乘以(sinπ/15 sin3π/15)/ (sinπ/15 sin3π/15)即可求。

Answer 3

第一步分子分母同除以2sin/π15，之后每一步同除以2，利用2倍角公式化简可得

Answer 4

cos(π/15) cos(2π/15) cos(3π/15) cos(4π/15) cos(5π/15) cos(6π/15) cos(7π/15)
=sin(8π/15)cos(3π/15)cos(5π/15)cos(7π/15) cos(6π/15)/(8sinπ/15)
=[sin14π/15/(16sinπ/15)]cos(3π/15)cos(5π/15)cos(6π/15)
=(1/16)4sin3π/15cos3π/15cos6π/15/[4sin(3π/15)] cos5π/15
=(1/16)*(sin3π/15)/(4sin3π/15)cos(5π/15)
=(1/64)*(1/2)=1/128