cosπ/5cos2π/5的值等于A.1/4; B.1/2; C2; D4(要演算过程)

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cosπ/5cos2π/5的值等于A.1/4;B.1/2;C2;D4(要演算过程)cosπ/5cos2π/5的值等于A.1/4;B.1/2;C2;D4(要演算过程)cosπ/5cos2π/5的值等于A

cosπ/5cos2π/5的值等于A.1/4; B.1/2; C2; D4(要演算过程)
cosπ/5cos2π/5的值等于
A.1/4; B.1/2; C2; D4
(要演算过程)

cosπ/5cos2π/5的值等于A.1/4; B.1/2; C2; D4(要演算过程)
cosπ/5cos2π/5
=2sinπ/5cosπ/5cos2π/5/(2sinπ/5)
=sin2π/5cos2π/5/(2sinπ/5)
=2sin2π/5cos2π/5/(4sinπ/5)
=sin4π/5/(4sinπ/5)
=sin(π-π/5)/(4sinπ/5)
=sinπ/5/(4sinπ/5)
=1/4
选A

先乘以sinπ/5,然后最后再除去就可以了,具体方法是:
cosπ/5cos2π/5
=(sinπ/5*cosπ/5cos2π/5)/(sinπ/5)
=sin2π/5cos2π/5/(2sinπ/5)
=sin4π/5/(4sinπ/5)
=sinπ/5/(4sinπ/5)
=1/4
选A,抄袭者不得好死~~

一个是sin54^一个是sin18^
一个小于二分之根号三,一个小于二分之一,所以小于二分之一
所以选A

原=(sin pi/5 * cos pi/5 * cos 2pi/5)/(sin pi/5)
=(sin 2pi/5 * cos 2pi/5)/(2sin pi/5)
=(sin 4pi/5)/(4sin pi/5)=1/4
注,添分母,配2倍角的正弦公式,利用互补.